Explanation:
Okay, so I need to figure out how many ways 7 people can sit around a round table, but with the condition that Pierre, Rosa, and Thomas (let me call them P, R, T for short) can't sit next to each other. And rotations are considered the same arrangement. Hmm, circular permutations can be tricky because rotations are considered identical, so usually we fix one person's position to avoid counting rotations multiple times. Let me recall the formula for circular permutations: for n people, it's (n-1)! because fixing one person and arranging the rest. But here, there are restrictions.

First, maybe I should find the total number of seating arrangements without any restrictions and then subtract the ones where at least two of P, R, T are sitting together. Wait, but inclusion-exclusion might be needed here because subtracting the cases where two are together might overcount or undercount. Alternatively, maybe it's better to use the principle of inclusion-exclusion. Let me think.

Alternatively, maybe I can approach it by first arranging the other people and then placing P, R, T in the gaps. That might be a better approach. Let me try that.

In circular permutations, arranging people around a table, if we want to ensure that certain people are not adjacent, sometimes we can use the gap method. Let me recall: if we have n people, the number of ways to arrange them around a table is (n-1)! If we want to place some people so that they are not adjacent, we can first arrange the others and then place the restricted people in the gaps.

So here, there are 7 people total. Let me consider the 3 people P, R, T who can't sit together. So maybe first arrange the other 4 people around the table, then place P, R, T in the gaps between them such that no two are adjacent.

But wait, in circular arrangements, arranging the 4 people first. Let me check: if we fix one person's position to account for rotational symmetry, then arrange the rest. So arranging 4 people around a table: (4-1)! = 6 ways. Then, between these 4 people, there are 4 gaps (since it's a circle). Wait, if there are 4 people around a table, there are 4 gaps between them. So if we want to place P, R, T into these gaps, each in separate gaps, so that no two are adjacent. Since there are 4 gaps and 3 people to place, the number of ways would be C(4,3) * 3! (choosing 3 gaps out of 4 and arranging the 3 people in them). So total arrangements would be 6 * C(4,3) * 3! = 6 * 4 * 6 = 144? Wait, but let me check again.

Wait, first arranging the 4 people: (4-1)! = 6. Then, the number of ways to place P, R, T into the gaps: since there are 4 gaps, and we need to choose 3 gaps, which is C(4,3) = 4, and then arrange P, R, T in those gaps, which is 3! = 6. So total is 6 * 4 * 6 = 144. But wait, the total number of circular permutations without restrictions is (7-1)! = 720. So 144 is less than 720, but is this the correct answer? Wait, but maybe I made a mistake here.

Wait, hold on. Let me think again. When arranging the 4 people around the table, fixing one person's position, so that rotations are accounted for. Then, the number of ways to arrange the other 3 people is 3! = 6? Wait, no. Wait, if we fix one person's position, then the remaining 3 people can be arranged in 3! ways. Wait, but the total number of arrangements for 4 people around a table is (4-1)! = 6. So that's correct. So arranging the 4 people: 6 ways. Then, inserting P, R, T into the gaps. Since there are 4 gaps between the 4 people, and we need to place 3 people into different gaps. So the number of ways is C(4,3) * 3! = 4 * 6 = 24. Then total arrangements would be 6 * 24 = 144. So that would be the answer? But let me check if this is correct.

Alternatively, maybe I need to consider that when arranging the 4 people, the number of gaps is 4, and inserting P, R, T into those gaps. But since the table is circular, inserting into gaps is okay. But let me think of another approach to verify.

Alternatively, total number of seating arrangements without restriction is 6! / 7? Wait no, circular permutations for n people is (n-1)! So for 7 people, it's 6! = 720. So if I use inclusion-exclusion, maybe I can compute the total number of arrangements where at least two of P, R, T are sitting together, and subtract that from 720.

But inclusion-exclusion can get complicated here. Let me try that approach as well to check.

First, total number of arrangements: 6! = 720.

Now, let me compute the number of arrangements where at least one pair of P, R, T are sitting together. Let me denote:

Let A be the set of arrangements where P and R are sitting together.

B be the set where P and T are sitting together.

C be the set where R and T are sitting together.

We need to compute |A ∪ B ∪ C|, and subtract that from total to get the desired number.

By inclusion-exclusion:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.

So first compute |A|, |B|, |C|. Since the problem is symmetric for P, R, T, each of these will be equal. So |A| = |B| = |C|.

Compute |A|: number of arrangements where P and R are sitting together. For circular tables, treating P and R as a single entity. But since it's circular, the number of arrangements is (n-1)! / n? Wait, no. Wait, when we have two people sitting together, we can consider them as a single "block". So for circular arrangements, the number of ways to arrange n-1 objects (the block and the other n-2 people) is (n-2)! But since the block can be arranged in 2 ways (P-R or R-P). Wait, but in circular permutations, when we have a block, how does that work?

Wait, let me recall. For linear arrangements, if we have two people together, we treat them as a block, so total n-1 elements, which can be arranged in (n-1)! ways, multiplied by 2 for the internal arrangement. But for circular arrangements, fixing one position to account for rotation. Let me think.

Alternatively, the formula for circular permutations with a block: if we have a block of k people, then the number of circular arrangements is (n - k)! * k! ? Wait, no. Wait, if we have a block, then effectively, we have (n - k + 1) objects to arrange? Wait, maybe not. Let me think again.

Suppose we have n people around a table. If we consider two people as a single block, then we have n - 1 objects to arrange (the block and the other n - 2 people). However, in circular permutations, the number of arrangements is (number of objects - 1)! So (n - 2)! And then multiply by 2 for the internal arrangement of the block. Therefore, |A| = 2 * (n - 2)! So for n = 7, |A| = 2 * 5! = 2 * 120 = 240. Wait, but hold on, total number of circular permutations is 6! = 720. If |A| is 240, then |A| + |B| + |C| = 3 * 240 = 720. But then when we subtract the intersections, etc., but let me check if this is correct.

Wait, but if we have 3 sets A, B, C each with 240 elements, then |A ∪ B ∪ C| would be 3*240 - ... but that seems like it might be overcounting. However, let me check with n=7.

Wait, but let me think again. Maybe my formula for |A| is wrong. Let me think of another way. For circular permutations, the number of ways where P and R are sitting together: fix P's position to remove rotational symmetry. Then R can be either to the left or right of P, so 2 choices. Then the remaining 5 people can be arranged in 5! ways. Wait, but if we fix P's position, then total arrangements are 6! / 7? Wait, no. Wait, fixing P's position, the total number of arrangements is 6! (since the rest can be arranged in 6! ways). Wait, but if we fix P's position, then the number of arrangements where R is adjacent to P is 2 * 5! (since R can be on the left or right, and the rest can be arranged in 5! ways). Therefore, the number of arrangements where P and R are adjacent is 2 * 5! = 240. But total number of arrangements is 6! = 720. So that seems correct. Therefore, |A| = |B| = |C| = 240. Then moving on.

Now, compute |A ∩ B|: arrangements where both P and R are together and P and T are together. Which implies that P, R, T are all sitting together. Because if P is adjacent to both R and T, then R and T are on either side of P, so all three are together. So the intersection of A and B is the set of arrangements where P, R, T are all sitting together. Similarly for other intersections.

So |A ∩ B| = |A ∩ C| = |B ∩ C| = number of arrangements where all three P, R, T are sitting together. Let me compute that.

Again, treating P, R, T as a single block. Then, we have 7 - 3 + 1 = 5 objects to arrange: the block and the other 4 people. For circular arrangements, the number of ways is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways (since P, R, T can be in any order within the block). Therefore, total is 24 * 6 = 144. Wait, but let me check with fixing positions again.

Alternatively, fix P's position. Then, since P, R, T are all together, we can think of the block as P, R, T in some order. If we fix P's position, then the block can be arranged in 2 ways: either R and T are on the left and right of P, or T and R are on the left and right. Wait, no. Wait, if we fix P's position, then the block can be arranged in 2! ways? Wait, no. Wait, if we fix P's position, then the block can be arranged in 2! ways? Wait, no. Wait, if the block is P, R, T, then once P is fixed, the other two can be arranged in 2! ways. But also, the block can be placed in two different directions? Wait, no. Wait, if we fix P's position, then the block can extend clockwise or counter-clockwise? Wait, no. Since it's a circular table, fixing P's position, the block can be arranged in 2! ways for the other two people. Wait, but actually, the entire block can be arranged in 3! ways, but since we fixed P's position, maybe?

Wait, this is getting confusing. Let me use the standard formula. If we have a block of 3 people, then the number of circular arrangements is (n - k)! * k! where n is total people, k is the block size? Wait, no. Wait, earlier, for a single block, we had (n - 1)! * 2 for two people. Wait, maybe not. Let me think again.

If we have a block of size k, then the number of circular arrangements is (n - k)! * k! ?

Wait, no. Wait, for linear arrangements, it's (n - k + 1)! * k! But for circular arrangements, it's different. Let me think of fixing one position. Suppose we fix one person's position to account for circular symmetry. Let's fix P's position. Then, the block of P, R, T can be arranged in 2! ways (since P is fixed, R and T can be arranged in 2! ways). But also, the block can be placed in different positions? Wait, no. If we fix P's position, then the block is fixed in terms of position? Wait, no. Wait, if we fix P's position, then the block can be arranged in two directions? Wait, maybe not. Let me think.

Alternatively, if we consider the block as a single entity, then with P fixed, the block can be placed in such a way that P is fixed, and the rest of the block is arranged. Wait, this is getting too tangled. Let me use the formula for circular permutations with a block. If we have a block of size k, then the number of circular arrangements is (n - k)! * k! ?

Wait, no. Wait, for example, if n = 3, and k = 3, then the number of circular arrangements is 1 (since all are together), but according to the formula (n - k)! * k! = 0! * 3! = 6, which is wrong. So that formula is incorrect.

Alternative approach: when considering circular permutations with a block, we can treat the block as a single entity. So, if we have n people, and we have a block of size k, then the number of circular arrangements is (n - k)! * k! / n? Wait, no. Wait, in linear arrangements, it's (n - k + 1)! * k! But for circular arrangements, it's different. Let me think of fixing one position to remove rotational symmetry.

Suppose we fix one person's position. Let's fix P's position. Then, the block of P, R, T can be arranged in 2! ways (since P is fixed, R and T can be arranged in 2! ways). Then, the remaining 4 people can be arranged in 4! ways. So total arrangements would be 2! * 4! = 2 * 24 = 48. But wait, but if we fix P's position, then the total number of arrangements where P, R, T are all together is 2! * 4! = 48. However, the total number of circular arrangements is 6! = 720. Wait, but if we fix P's position, the total number of arrangements is 6! / 7? No, fixing P's position, the total number of arrangements is 6! / 7? Wait, no. Wait, fixing one person's position in circular permutations, the number of arrangements is (n - 1)! So for n = 7, fixing P's position, the total number of arrangements is 6! = 720. Wait, no! Wait, no. Wait, if you fix one person's position, the number of arrangements is (n - 1)! So for n = 7, fixing P's position, the number of arrangements is 6! = 720? Wait, no! Wait, no. Wait, if you fix one person's position, the number of arrangements is (n - 1)! So for n = 7, fixing one person, the number of arrangements is 6! = 720? Wait, no! Wait, no. Wait, (n - 1)! is the total number of circular permutations. If you fix one person's position, then the rest can be arranged in (n - 1)! / n? No, no. Wait, confusion here.

Let me recall: the number of circular permutations of n people is (n - 1)! because rotations are considered the same. If we fix one person's position, then the number of arrangements is (n - 1)! / n * n? Wait, no. Wait, fixing one person's position effectively removes rotational symmetry, so the number of arrangements is (n - 1)! So for example, for n = 3, fixing one person, the other two can be arranged in 2! ways, which is (3 - 1)! = 2. Correct. So for n = 7, fixing one person's position, the number of arrangements is 6! = 720? Wait, no! Wait, (n - 1)! is 6! = 720. But if you fix one person's position, then the number of arrangements is (n - 1)! So for n = 7, fixing one person, the number of arrangements is 6! = 720? Wait, that can't be. Wait, no. Wait, if you fix one person's position, then the number of arrangements is (n - 1)! So for n = 7, it's 6! = 720? Wait, but that would mean that the total number of circular permutations is 720, which is correct. So if we fix one person's position, the number of arrangements is 6! = 720. Wait, but that seems contradictory. Wait, no. Wait, the total number of circular permutations is (n - 1)! So for n = 7, it's 6! = 720. If we fix one person's position, then the number of arrangements is (n - 1)! So that is the same as the total number of circular permutations. Wait, that seems redundant. Wait, no. Wait, fixing one person's position is a way to count circular permutations. So if you fix one person's position, then the rest can be arranged in (n - 1)! ways. Wait, no. Wait, if you fix one person's position, then the number of ways to arrange the rest is (n - 1)! / n? No. Wait, confusion arises here. Let me check with n = 3. If you fix one person's position, then the other two can be arranged in 2! ways, which is (3 - 1)! = 2. So yes, fixing one person's position, the number of arrangements is (n - 1)! So for n = 7, fixing one person's position, the number of arrangements is 6! = 720. Wait, but that's the same as the total number of circular permutations. Wait, but that's because fixing one person's position is a way to count them. So if you fix one person's position, you are effectively counting the circular permutations as linear permutations of the remaining people. So, okay, that seems correct.

So, going back, if we fix P's position, then the number of arrangements where P, R, T are all together is calculated as follows: since P is fixed, the block of P, R, T can be arranged in 2! ways (since P is fixed, R and T can be on either side). Wait, but actually, the block can be arranged in 2! ways? Wait, if P is fixed, then the block can be P-R-T or P-T-R? Wait, but since it's a circular table, once P is fixed, the block can be arranged in two different ways: R to the left of P and T to the right, or T to the left and R to the right. So that's 2 ways. Then, the remaining 4 people can be arranged in 4! ways. Therefore, total arrangements where P, R, T are all together is 2 * 4! = 2 * 24 = 48. Therefore, the number of circular arrangements where P, R, T are all together is 48. But wait, but earlier I thought it was 144. Which is correct?

Wait, if we don't fix positions, then the number of circular arrangements where P, R, T are all together is calculated by treating them as a block. So, number of objects is 5 (the block + 4 others). Number of circular arrangements is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways. So total is 24 * 6 = 144. But if we fix P's position, we get 48. Which is different. So which is correct?

Wait, if we fix P's position, then the total number of arrangements is 6! = 720. But if we don't fix positions, the total number of circular arrangements is 6! = 720. So if we fix P's position, the number of arrangements where P, R, T are all together is 48. Therefore, the total number without fixing positions would be 48 * 7 / 1? Wait, no. Wait, if we fix P's position, we are considering only 1/n of the total arrangements? Wait, no. Wait, fixing P's position is a way to count all circular arrangements. So if we fix P's position, the number of arrangements where P, R, T are all together is 48, which would be the same as the total number of circular arrangements where P, R, T are all together. Wait, but according to the other method, treating the block as a single entity, we get 144. Which is different. So which is correct?

Wait, let me think. If we don't fix positions, the number of circular arrangements where P, R, T are all together is calculated as follows: consider the block of P, R, T as a single entity. Then, we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways. So total is 24 * 6 = 144. However, if we fix P's position, we get 48. But 144 is 3 times 48. Why?

Ah! Because when we fix P's position, we are considering only arrangements where P is fixed. However, in the total number of circular arrangements where P, R, T are all together, P can be in any position. So if we fix P's position, we are counting only the arrangements where P is fixed, but in reality, there are multiple positions where P can be. Wait, but in circular permutations, fixing P's position accounts for all rotations. Therefore, if we fix P's position, the number of arrangements where P, R, T are all together is 48, which should be equal to the total number of such arrangements divided by 7 (since there are 7 positions for P). Wait, but if total number is 144, then 144 / 7 is not an integer. Which is a problem. Therefore, one of the methods is wrong.

Wait, this suggests that there is a mistake in one of the approaches. Let me check again.

First approach: treating the block as a single entity. Number of circular arrangements is (5 - 1)! * 3! = 24 * 6 = 144. But if we fix P's position, we get 48. Let me check with n=3. If n=3, and we want the number of circular arrangements where all 3 are together. Well, there's only 1 arrangement, since they are all together. But according to the first method: treating the block as a single entity, number of circular arrangements is (1 - 1)! * 3! = 0! * 6 = 1 * 6 = 6. Which is wrong. So clearly, that formula is incorrect. Therefore, my previous approach was wrong.

So, the formula for circular permutations with a block is not (n - k)! * k! So that approach is invalid. Therefore, need to think differently.

Alternative approach: For circular arrangements, the number of ways where a specific set of people are all together can be calculated by considering the block as a single entity. However, in circular permutations, when you have a block, the formula is different. Let me check.

In linear arrangements, the number of ways to arrange n people with a block of k people together is (n - k + 1)! * k!.

But for circular arrangements, it's a bit different. Let me think. If we have a block of k people, then the number of circular arrangements is (n - k)! * k! / n? Wait, not sure. Wait, here's a method: fix the block's position to account for circular symmetry. For example, fix the block in a specific position, then arrange the rest. But since the table is circular, fixing the block's position removes rotational symmetry.

Wait, suppose we fix the block of P, R, T in a specific position. Then, the number of ways to arrange the block is 3! (since they can be in any order). Then, the remaining 4 people can be arranged in 4! ways. However, since we fixed the block's position, we don't have to worry about rotations. Therefore, total number of arrangements is 3! * 4! = 6 * 24 = 144. But this seems to contradict the previous result when fixing P's position. Wait, but if we fix the block's position, then we are not accounting for rotations. However, in circular permutations, fixing the block's position is equivalent to considering rotational symmetry. Therefore, the total number of circular arrangements where P, R, T are all together is 3! * 4! = 144. However, when we fixed P's position earlier, we got 48. Which is different. So why?

Because when we fixed P's position, we were considering only arrangements where P is fixed, but in reality, the block can be anywhere. However, if we fix P's position, then the block can be arranged in two different ways (as discussed earlier), and the rest can be arranged in 4! ways. So 2 * 4! = 48. But according to the other method, it's 144. Which is 3 times larger. Where is the discrepancy?

Ah! Because when we fix P's position, we are considering only arrangements where P is fixed, but in reality, the block can be placed in different positions relative to P. Wait, no. If we fix P's position, then the block must include P. So the block is fixed with P in a specific position, and the rest of the block (R and T) can be arranged in 2 ways. Then, the rest of the people can be arranged in 4! ways. Therefore, total is 2 * 4! = 48. However, if we don't fix P's position, the total number of arrangements where P, R, T are all together is 144. Therefore, 144 is equal to 7 * 48 / something? Wait, no. Wait, if we fix P's position, we are considering 1/7 of all possible arrangements. Therefore, if the total number of arrangements where P, R, T are all together is 144, then fixing P's position would give 144 / 7? But 144 / 7 is not an integer. Which is a problem. Therefore, there must be an error in one of the methods.

Wait, let me think again. Let me use the formula for circular permutations with a block. Suppose we have n people, and we want to count the number of circular arrangements where a specific set of k people are all together. The formula is (n - k)! * k! / n? Wait, no. Wait, here's a resource I remember: when arranging people around a circular table with a group of people together, the number of arrangements is (n - k)! * k! / n? Wait, not sure. Wait, let me think of it as follows.

In circular permutations, the number of ways to arrange n people is (n - 1)! If we want to have a specific group of k people together, we can consider them as a single entity. Then, we have (n - k + 1) entities to arrange. However, in circular permutations, the number of ways to arrange (n - k + 1) entities is (n - k)! So total number of arrangements is (n - k)! * k! (for arranging the group internally). But wait, for example, with n = 3, k = 3: (3 - 3)! * 3! = 1 * 6 = 6, which is wrong because there's only 1 circular arrangement. So that formula is wrong.

Wait, maybe the formula is (n - k)! * k! / n? For n = 3, k = 3: (0)! * 6 / 3 = 1 * 6 / 3 = 2, which is still wrong. Hmm.

Alternative approach: Let me think of the number of ways where P, R, T are all together. First, consider them as a single block. Then, we have 5 entities: the block and the other 4 people. The number of circular arrangements of these 5 entities is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways. So total is 24 * 6 = 144. However, in this calculation, are we overcounting? For example, if the block is considered as a single entity, but in circular arrangements, rotating the entire block doesn't create a new arrangement. Wait, but since we already accounted for circular arrangements by using (5 - 1)! for the entities, then multiplying by the internal arrangements of the block should be correct. However, when n = 3, this would give (3 - 3)! * 3! = 1 * 6 = 6, which is wrong. But for n = 3, the correct number is 1. So why is there a discrepancy?

Ah! Because when the block is the entire set of people, the formula (n - k)! * k! counts linear arrangements, not circular. Wait, no. Wait, if we have n = 3, k = 3, then treating them as a single block, we have 1 entity. The number of circular arrangements of 1 entity is 1 (since there's only one way to arrange a single entity around a table). Then, multiply by k! = 6, which would give 6, which is wrong. Therefore, clearly, this approach is incorrect.

Therefore, there must be a different way. Let me think again. Maybe the formula for circular permutations with a block is different. Let me refer to standard results.

Upon recalling, the number of ways to arrange n people around a circular table with a specific group of k people sitting together is (n - k)! * k! / n? Wait, no. Wait, here's a better way. Let me fix the position of the block to eliminate rotational symmetry. Suppose we fix the block in a specific position. Then, the number of ways to arrange the block is k! and the rest can be arranged in (n - k)! ways. However, since we fixed the block's position, we don't need to divide by anything. Therefore, total number of arrangements is k! * (n - k)! But for n = 3, k = 3, this would be 6 * 1 = 6, which is still wrong. But the correct number is 1. So this is not correct.

Wait, but if we fix the block's position, then the number of circular arrangements is (n - k)! * k! / n? No. Wait, confusion reigns. Let me try with n = 3, k = 2. Suppose we want the number of circular arrangements where two specific people are together. Let's say P and R. Fix P's position. Then R can be to the left or right of P, so 2 choices. The third person, T, can be arranged in 1! way. So total is 2 * 1 = 2. But the total number of circular arrangements is 2! = 2. Wait, but if we don't fix positions, the number of circular arrangements where P and R are together is 2. Which is correct. However, according to the formula (n - k)! * k! = (3 - 2)! * 2! = 1 * 2 = 2, which matches. Wait, but earlier with n = 3, k = 3, it would be (3 - 3)! * 3! = 1 * 6 = 6, which is wrong. But if we fix the block's position, then for n = 3, k = 3, fixing the block's position, we have 1 way to arrange the block (since it's the entire table), and the rest is 0! = 1, so total is 1 * 6 = 6? But that's wrong. Wait, but if we fix the block's position, then the number of arrangements is k! * (n - k)! = 6 * 1 = 6, but in reality, there's only 1 circular arrangement. So why is there a discrepancy?

Ah! Because when we fix the block's position, we are not accounting for rotational symmetry. Wait, if we fix the block's position, then we are effectively making it a linear arrangement. Wait, no. If we fix the block's position on the table, then the rest of the people are arranged relative to it. However, in circular permutations, fixing the block's position removes rotational symmetry. Therefore, the number of arrangements should be k! * (n - k)! But for n = 3, k = 3, this would be 6 * 1 = 6, but there's only 1 circular arrangement. Therefore, this formula is wrong.

Wait, but if we fix the block's position, then the number of arrangements is k! * (n - k)! / something? Wait, no. Wait, if we fix the block's position, then the number of arrangements is k! * (n - k)! / n? No. Wait, confusion. Let me think of n = 3, k = 2. If we fix the block of P and R, then the third person T can be arranged in 1! ways. But since we fixed the block's position, the total number of arrangements is 2! * 1! = 2, which matches the correct number. However, if we don't fix the block's position, the number of circular arrangements where P and R are together is 2, which is equal to (n - k)! * k! = 1! * 2! = 2. So that works. But for n = 3, k = 3, it would be (n - k)! * k! = 0! * 3! = 6, which is wrong. But if we fix the block's position, we get 3! * 0! = 6, which is wrong. So why does it work for k = 2 but not for k = 3?

Ah! Because when k = n, the formula (n - k)! * k! = 0! * n! = n! which is the total number of circular permutations, but in reality, there is only 1 circular permutation. So clearly, the formula is not correct for k = n. Therefore, there must be a different formula.

Wait, but when k = n, the number of circular arrangements where all people are together is 1 (since they are all together). But according to the formula (n - k)! * k! = n! which is wrong. Therefore, the formula is only valid when k < n? Or maybe there's a different approach.

Wait, maybe the formula is (n - k)! * k! / n? For n = 3, k = 3: 0! * 6 / 3 = 2, which is still wrong. For n = 3, k = 2: 1! * 2 / 3 = 2/3, which is wrong. So that doesn't work.

This is getting too confusing. Maybe I should go back to the original problem and try to use the first method where I arranged the other people first and then placed P, R, T in the gaps. Earlier, I calculated that as 6 * 4 * 6 = 144. But if the total number of circular permutations is 720, then 144 is 1/5 of that. But let me check with another approach.

Alternatively, let me think of the total number of ways where no two of P, R, T are adjacent. This is equal to total number of circular permutations minus the number of permutations where at least two are adjacent. But earlier, I tried inclusion-exclusion but got stuck on calculating |A ∩ B|.

Alternatively, let me try to use the gap method again. Let me try to calculate it properly.

First, arrange the other 4 people around the table. Since it's a circular table, the number of ways is (4 - 1)! = 6. Now, between these 4 people, there are 4 gaps. We need to place P, R, T into these gaps such that no two are in the same gap (to ensure they are not adjacent). The number of ways to choose 3 gaps out of 4 is C(4, 3) = 4. Then, arrange P, R, T in these gaps: 3! = 6. So total number of ways is 6 * 4 * 6 = 144. So this gives 144. But earlier, when I tried inclusion-exclusion, I was confused about the |A ∩ B| term. Let me try to proceed with inclusion-exclusion now, using this result.

If the answer is 144, then the total number of valid arrangements is 144. But let me check with inclusion-exclusion.

Total number of circular permutations: 6! = 720.

Number of arrangements where at least one pair of P, R, T are adjacent: |A ∪ B ∪ C|.

As before, |A| = |B| = |C| = 2 * 5! = 240. Wait, earlier I thought that fixing P's position gives 2 * 5! = 240, but that seems high. Wait, if total number of circular permutations is 720, and |A| is 240, then |A| + |B| + |C| = 720. Then, subtracting intersections, etc. But if |A ∩ B| is the number of arrangements where P and R are together AND P and T are together, which implies all three are together. So |A ∩ B| = number of arrangements where all three are together. Let me calculate that.

If all three are together, then as per the gap method, if we consider them as a block, then number of circular arrangements is (5 - 1)! * 3! = 24 * 6 = 144? Wait, but earlier we saw that this might be wrong. But if we use the gap method, we have 144 as the number of valid arrangements, which would mean that the number of invalid arrangements is 720 - 144 = 576. But according to inclusion-exclusion, if |A ∪ B ∪ C| = 3*240 - 3*|A ∩ B| + |A ∩ B ∩ C|. Let me assume that |A ∩ B| = number of arrangements where all three are together, which we need to calculate.

If we assume that |A ∩ B| = number of arrangements where all three are together, which is 144, then |A ∩ B| = |A ∩ C| = |B ∩ C| = 144. Then, |A ∩ B ∩ C| is also the number of arrangements where all three are together, which is 144. Wait, but that can't be. Because |A ∩ B ∩ C| is the set of arrangements where all three pairs are together, which is the same as all three being together. So |A ∩ B ∩ C| = 144.

Therefore, |A ∪ B ∪ C| = 3*240 - 3*144 + 144 = 720 - 432 + 144 = 720 - 432 = 288 + 144 = 432? Wait, 3*240 = 720, then subtract 3*144 = 432, so 720 - 432 = 288, then add back 144, so 288 + 144 = 432. Therefore, |A ∪ B ∪ C| = 432. Therefore, the number of valid arrangements is total - invalid = 720 - 432 = 288. But this contradicts the gap method result of 144. So which is correct?

This suggests that there is a mistake in one of the methods. Let me check with smaller numbers. Let me take n = 4, with 3 people who can't sit together. Wait, maybe that's too small. Let me try n = 4, with 3 people P, R, T and one other person, say Q. How many circular arrangements where P, R, T are not all adjacent? Wait, but in n = 4, if we have 3 people who can't sit together... Wait, in a circular table of 4 people, if we want to ensure that P, R, T are not all adjacent. Wait, but in a table of 4, any three people will be adjacent if they are seated together. Wait, no. In a circular table of 4, if three people are seated, they are adjacent if they are next to each other. But with 4 people, the number of ways to seat them is (4-1)! = 6. How many of these have P, R, T all together? Let's calculate.

Using the gap method: arrange the other person Q around the table. Since there's only one person, arranging Q around the table: (1-1)! = 1 way. Then, there are 1 gap between Q. Wait, but we need to place P, R, T into gaps. But there's only 1 gap, and we need to place 3 people. Which is impossible. Therefore, the number of valid arrangements is 0? But that can't be. Wait, in a circular table of 4 people, if we have 3 people who cannot sit together, but since there are only 4 seats, and 3 people, they must be seated with one seat between them? Wait, no. Wait, if you have 4 seats, and 3 people, then one seat is empty. The three people can be seated such that no two are adjacent? Wait, in a circular table of 4 seats, can 3 people be seated with no two adjacent? No, because if you have 3 people, each person has two neighbors. In a circle of 4, if you seat 3 people, at least two of them must be adjacent. Because with 3 people, there are 3 seats occupied and 1 empty. The empty seat is between two people, so those two people are adjacent. Therefore, it's impossible to seat 3 people around a table of 4 without having at least two adjacent. Therefore, the number of valid arrangements is 0. Which matches the gap method result. But according to inclusion-exclusion, let's see:

Total number of circular permutations: 6.

Number of arrangements where at least one pair of P, R, T are adjacent. Let's compute |A ∪ B ∪ C|.

Each |A|, |B|, |C|: number of arrangements where a specific pair is together. For n = 4, fixing one person's position, say Q. Then, the number of arrangements where P and R are together: treat P and R as a block. Then, we have Q and the block. Number of circular arrangements: (2 - 1)! = 1! = 1. Multiply by 2 (for P-R or R-P): 2. Then, the total number of arrangements where P and R are together is 2. Similarly for |B| and |C|. So |A| + |B| + |C| = 3 * 2 = 6. Now, |A ∩ B|: arrangements where P and R are together AND P and T are together, which means all three are together. Number of such arrangements: treat P, R, T as a block. Then, we have Q and the block. Number of circular arrangements: (2 - 1)! = 1. Multiply by 3! = 6 for internal arrangements. So total is 6. Therefore, |A ∩ B| = |A ∩ C| = |B ∩ C| = 6. Then, |A ∩ B ∩ C| is also 6. Therefore, |A ∪ B ∪ C| = 6 - 3*6 + 6 = 6 - 18 + 6 = -6, which is impossible. Clearly, something is wrong here.

This suggests that my inclusion-exclusion approach is flawed. Therefore, there must be a mistake in how I'm calculating |A|, |B|, |C|, etc. For n = 4, the total number of circular permutations is 6. If I calculate |A| as 2, then 3*2 = 6, which equals the total number of permutations. Therefore, |A ∪ B ∪ C| = 6, so the number of valid arrangements is 6 - 6 = 0, which is correct. But according to the inclusion-exclusion formula, I had |A ∪ B ∪ C| = 3*2 - 3*6 + 6 = -6, which is wrong. Therefore, my calculation of |A ∩ B| was incorrect.

So, what is the correct value of |A ∩ B|? For n = 4, |A ∩ B| is the number of arrangements where both P and R are together and P and T are together, which means all three are together. How many such arrangements are there? If we treat P, R, T as a block, then we have the block and Q. Number of circular arrangements: (2 - 1)! = 1. Multiply by the number of internal arrangements of the block: 3! = 6. So total is 6. But the total number of circular permutations is 6, so this suggests that all arrangements have all three together, which is not true. Wait, but in reality, when n = 4, any arrangement of P, R, T, Q must have at least two of P, R, T together. Wait, but if you have 3 people and 1 empty seat, then in any arrangement, at least two of the three people are adjacent. Therefore, all arrangements have at least one pair together, so |A ∪ B ∪ C| = 6, and the number of valid arrangements is 0. But according to the inclusion-exclusion formula, if |A| = 2, |B| = 2, |C| = 2, |A ∩ B| = 6, |A ∩ C| = 6, |B ∩ C| = 6, and |A ∩ B ∩ C| = 6, then:

|A ∪ B ∪ C| = 3*2 - 3*6 + 6 = 6 - 18 + 6 = -6, which is impossible. Therefore, clearly, my calculation of |A ∩ B| is wrong.

Wait, what is the correct value of |A ∩ B|? For n = 4, |A ∩ B| is the number of arrangements where both P and R are together and P and T are together. Which means all three are together. How many such arrangements are there? Let's list them. Fix Q's position. Then, the block of P, R, T can be arranged in 2 ways: clockwise or counter-clockwise? Wait, no. If Q is fixed, then the block can be arranged in 2 ways: P-R-T or P-T-R, but since it's a circular table, fixing Q's position, the block can be placed in two different ways? Wait, no. If Q is fixed, then the block of P, R, T occupies the other three seats. Since it's a circular table, the block can be arranged in 2 ways: clockwise or counter-clockwise? Wait, no. If Q is fixed, then the three seats adjacent to Q are in a line? No, in a circular table with Q fixed, the other three seats are arranged around Q. If we have a block of P, R, T, then they can be arranged in 3! = 6 ways. But since the table is circular, but Q is fixed, so rotations are already accounted for. Therefore, the number of arrangements where P, R, T are all together is 3! = 6. But the total number of circular permutations is 6, so this suggests that all arrangements have P, R, T together, which is true, because with 3 people and 1 empty seat, they must be adjacent. Therefore, |A ∩ B| = 6. But then, in inclusion-exclusion, we have:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| = 2 + 2 + 2 - 6 - 6 - 6 + 6 = 6 - 18 + 6 = -6, which is impossible. Therefore, clearly, my calculation of |A| is wrong.

Wait, earlier I said |A| = 2 for n = 4. Let me recalculate |A|. For n = 4, |A| is the number of arrangements where P and R are together. Fix Q's position. Then, P and R can be seated together in two ways: either on the left and right of Q, or on the other side? Wait, no. If Q is fixed, there are two seats adjacent to Q. Let me visualize the table: Q is fixed at position 1. The other seats are 2, 3, 4. Seat 2 is adjacent to Q (seat 1), seat 4 is adjacent to Q. Seat 3 is opposite Q. If P and R are to be seated together, they can be in seats 2 and 3, or seats 3 and 4, or seats 4 and 2? Wait, no. If P and R are to be adjacent, they need to be in seats that are next to each other. In a circular table with Q fixed at seat 1, the adjacent seats are 2 and 4. Seat 3 is opposite. So the possible adjacent pairs are (2,3), (3,4), (4,2). Wait, but seat 3 is not adjacent to seat 2 or 4? No, in a circular table of 4 seats, each seat is adjacent to two others. So seat 1 (Q) is adjacent to 2 and 4. Seat 2 is adjacent to 1 and 3. Seat 3 is adjacent to 2 and 4. Seat 4 is adjacent to 3 and 1. So the adjacent pairs are (1,2), (2,3), (3,4), (4,1). So if Q is fixed at seat 1, then the adjacent pairs involving Q are (1,2) and (1,4). The other adjacent pairs are (2,3) and (3,4). So if we want P and R to be adjacent, they can be in (2,3), (3,4), or (1,2) or (1,4). But since Q is fixed at seat 1, if P and R are in (1,2), then one of them is Q? No, Q is fixed at seat 1, so P and R cannot be in seat 1. Therefore, P and R can only be in (2,3) or (3,4). So two possible adjacent pairs. For each of these, P and R can be arranged in 2 ways. Then, the remaining seat (seat 4 or seat 2) is occupied by T. So total number of arrangements where P and R are together is 2 (adjacent pairs) * 2 (arrangements of P and R) * 1 (T in the remaining seat) = 4. But earlier I thought it was 2. So my previous calculation was wrong. Therefore, |A| = 4, not 2. Similarly for |B| and |C|. So |A| + |B| + |C| = 3 * 4 = 12. But the total number of circular permutations is 6, which is impossible. Wait, this suggests another error.

Wait, no. If Q is fixed, the total number of arrangements is 3! = 6. If |A| is the number of arrangements where P and R are together, which we calculated as 4, but that's more than the total number of arrangements. Clearly, something is wrong.

Wait, let me list all possible arrangements with Q fixed at seat 1. The other seats are 2, 3, 4. The people are P, R, T. The possible arrangements are:

1. P, R, T in seats 2, 3, 4: P in 2, R in 3, T in 4

2. P, R, T in seats 2, 3, 4: P in 2, T in 3, R in 4

3. P, R, T in seats 2, 3, 4: R in 2, P in 3, T in 4

4. P, R, T in seats 2, 3, 4: R in 2, T in 3, P in 4

5. P, R, T in seats 2, 3, 4: T in 2, P in 3, R in 4

6. P, R, T in seats 2, 3, 4: T in 2, R in 3, P in 4

Now, which of these have P and R adjacent? Let's check:

1. P in 2, R in 3: adjacent (seats 2 and 3 are adjacent). Yes.

2. P in 2, T in 3, R in 4: P and R are not adjacent (seat 2 and 4 are not adjacent). No.

3. R in 2, P in 3: adjacent. Yes.

4. R in 2, T in 3, P in 4: R and P are not adjacent. No.

5. T in 2, P in 3, R in 4: P and R are not adjacent. No.

6. T in 2, R in 3, P in 4: R and P are not adjacent. No.

So only arrangements 1 and 3 have P and R adjacent. So |A| = 2. Similarly, |B| (P and T adjacent) and |C| (R and T adjacent) would also be 2 each. So total |A| + |B| + |C| = 6. Now, |A ∩ B| is the number of arrangements where P and R are together AND P and T are together, which means all three are together. How many such arrangements are there? From the list above, arrangements 1 and 3 have P and R together. Let's check if they also have P and T together.

Arrangement 1: P in 2, R in 3, T in 4. P is adjacent to R (seats 2-3) and T is in seat 4, which is adjacent to seat 3 (R) and seat 1 (Q). So P is not adjacent to T. So no.

Arrangement 3: R in 2, P in 3, T in 4. P is adjacent to R (seats 2-3) and T is in seat 4, adjacent to P (seat 3-4). So P is adjacent to T. So this arrangement has P adjacent to both R and T. So only arrangement 3 has all three together. Similarly, if we check for other intersections, like |A ∩ C| (P and R together and R and T together), which would also require all three together. Let's see: in arrangement 3, R is adjacent to P and T, so it's included. Are there others? Let's check arrangement 6: T in 2, R in 3, P in 4. R is adjacent to T (seats 2-3) and P is in seat 4, adjacent to R (seat 3-4). So this arrangement also has R adjacent to both T and P. So this is another arrangement where all three are together. Wait, but in our list above, arrangement 6 is T in 2, R in 3, P in 4. So R is between T and P, so all three are adjacent. So this is another arrangement where all three are together. Similarly, arrangement 5: T in 2, P in 3, R in 4. P is adjacent to T (seats 2-3) and R is in seat 4, adjacent to P (seat 3-4). So this also has all three together. Wait, but earlier I thought only arrangement 3 had all three together. Let me check:

Arrangement 1: P, R, T in seats 2, 3, 4: P in 2, R in 3, T in 4. P is adjacent to R, R is adjacent to T, but P is not adjacent to T. So only two adjacent pairs.

Arrangement 2: P in 2, T in 3, R in 4: P and T are adjacent (seats 2-3), T and R are adjacent (seats 3-4), but P and R are not adjacent.

Arrangement 3: R in 2, P in 3, T in 4: R and P are adjacent, P and T are adjacent, R and T are not adjacent.

Arrangement 4: R in 2, T in 3, P in 4: R and T are adjacent, T and P are adjacent, R and P are not adjacent.

Arrangement 5: T in 2, P in 3, R in 4: T and P are adjacent, P and R are adjacent, T and R are not adjacent.

Arrangement 6: T in 2, R in 3, P in 4: T and R are adjacent, R and P are adjacent, T and P are not adjacent.

So actually, there are 3 arrangements where all three are together: arrangements 3, 5, 6? Wait, no. Wait, arrangement 3: R, P, T in seats 2, 3, 4. P is between R and T, so all three are adjacent. Similarly, arrangement 5: T, P, R in seats 2, 3, 4. P is between T and R, so all three are adjacent. Arrangement 6: T, R, P in seats 2, 3, 4. R is between T and P, so all three are adjacent. So there are 3 arrangements where all three are together. Therefore, |A ∩ B| = |A ∩ C| = |B ∩ C| = 3? Wait, no. Let me think.

|A ∩ B| is the number of arrangements where P and R are together AND P and T are together. From the list, only arrangement 3 has P and R together and P and T together. Similarly, |A ∩ C| is the number of arrangements where P and R are together AND R and T are together. Which would be arrangements 3 and 6? Wait, arrangement 3 has P and R together and R and T not together? No, in arrangement 3, R is in 2, P in 3, T in 4. R is adjacent to P (seats 2-3), and T is in 4, adjacent to P (seats 3-4). So R is not adjacent to T. So only P and R are together, and P and T are together. So arrangement 3 has both P and R together and P and T together. Similarly, arrangement 5: T in 2, P in 3, R in 4. P is adjacent to T and R, so this has P and T together and P and R together. So this is also in |A ∩ B|. Similarly, arrangement 6: T in 2, R in 3, P in 4. R is adjacent to T and P, so this is in |B ∩ C|. So |A ∩ B| includes arrangements 3 and 5? Wait, no. Let me clarify:

|A ∩ B| is arrangements where P and R are together AND P and T are together. So in arrangement 3: P and R are together, P and T are together. In arrangement 5: P and T are together, P and R are together. So both arrangements 3 and 5 are in |A ∩ B|. Similarly, arrangement 6 is in |B ∩ C| (R and T together, R and P together). So |A ∩ B| has 2 arrangements, |A ∩ C| has 2 arrangements, |B ∩ C| has 2 arrangements. And |A ∩ B ∩ C| is the number of arrangements where all three pairs are together, which is all three together. How many such arrangements are there? From the list, arrangements 3, 5, 6 have all three together? No, in arrangement 3, P is between R and T, so all three are together. In arrangement 5, P is between T and R, so all three are together. In arrangement 6, R is between T and P, so all three are together. So there are 3 arrangements where all three are together. Therefore, |A ∩ B ∩ C| = 3.

So now, applying inclusion-exclusion:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| = 2 + 2 + 2 - 2 - 2 - 2 + 3 = 6 - 6 + 3 = 3. But the total number of arrangements is 6, so the number of valid arrangements is 6 - 3 = 3, which is wrong because we know it should be 0. Wait, this is still inconsistent. What's the issue?

Wait, the valid arrangements are those where no two of P, R, T are adjacent. But in n = 4, it's impossible, so the number should be 0. But according to this, it's 3. Clearly, something is wrong. Let me check the calculations again.

|A| = 2, |B| = 2, |C| = 2. Total 6.

|A ∩ B| = 2 (arrangements 3 and 5), |A ∩ C| = 2 (arrangements 3 and 6?), |B ∩ C| = 2 (arrangements 6 and 5?).

Wait, let's define:

|A ∩ B|: arrangements where P and R are together AND P and T are together. These are arrangements where all three are together. From the list, arrangements 3 and 5 have P and R together and P and T together. Wait, arrangement 3: P is between R and T, so P is adjacent to both. Arrangement 5: P is between T and R, same thing. So |A ∩ B| = 2.

Similarly, |A ∩ C|: arrangements where P and R are together AND R and T are together. These are arrangements where all three are together. From the list, arrangement 3: R is between P and T, so yes. Arrangement 6: R is between T and P, so yes. So |A ∩ C| = 2.

Similarly, |B ∩ C|: arrangements where P and T are together AND R and T are together. These are arrangements where all three are together. From the list, arrangement 5: T is between P and R, and arrangement 6: T is between R and P. Wait, arrangement 5: T in 2, P in 3, R in 4. T is adjacent to P and R. Arrangement 6: T in 2, R in 3, P in 4. T is adjacent to R and P. So |B ∩ C| = 2.

Therefore, |A ∩ B| = |A ∩ C| = |B ∩ C| = 2.

And |A ∩ B ∩ C| is the number of arrangements where all three pairs are together, which is all three together. From the list, there are 3 such arrangements: 3, 5, 6. Wait, but earlier I thought |A ∩ B| = 2, but if |A ∩ B ∩ C| is 3, then how?

Wait, no. |A ∩ B ∩ C| is the set of arrangements that are in all three intersections, i.e., where P and R are together, P and T are together, and R and T are together. Which is the same as all three being together. How many such arrangements are there? In the list, arrangements 3, 5, 6 all have all three together. So |A ∩ B ∩ C| = 3.

Therefore, applying inclusion-exclusion:

|A ∪ B ∪ C| = 2 + 2 + 2 - 2 - 2 - 2 + 3 = 6 - 6 + 3 = 3. But the total number of arrangements is 6, so the number of valid arrangements is 6 - 3 = 3, which is wrong. But we know it should be 0. Where is the mistake?

Ah! The mistake is in the calculation of |A|, |B|, |C|. Earlier, I said |A| = 2, but actually, in the list, there are 2 arrangements where P and R are together, but there are also arrangements where P and R are not together. But the total number of arrangements is 6, and |A ∪ B ∪ C| = 3, which would mean that 3 arrangements have at least one pair together, and 3 have none. But we know that in n = 4, it's impossible to have none. So clearly, there is a mistake in the calculation of |A|, |B|, |C|.

Wait, let me re-examine |A|. |A| is the number of arrangements where P and R are together. From the list, arrangements 1 and 3 have P and R together. So |A| = 2. Similarly, |B| (P and T together) is arrangements 2, 5. Wait, arrangement 2: P in 2, T in 3, R in 4. P and T are adjacent. Arrangement 5: T in 2, P in 3, R in 4. P and T are adjacent. So |B| = 2. Similarly, |C| (R and T together) is arrangements 2, 4, 6? Wait, arrangement 2: R in 4, T in 3, so R and T are adjacent. Arrangement 4: R in 2, T in 3, so R and T are adjacent. Arrangement 6: R in 3, T in 2, so R and T are adjacent. So |C| = 3? Wait, no. Let me check:

Arrangement 1: P, R, T in seats 2, 3, 4: P in 2, R in 3, T in 4. R and T are adjacent (seats 3-4). So this is in |C|.

Arrangement 2: P in 2, T in 3, R in 4: R and T are adjacent (seats 3-4). So in |C|.

Arrangement 3: R in 2, P in 3, T in 4: R and T are not adjacent (seats 2 and 4 are not adjacent). So not in |C|.

Arrangement 4: R in 2, T in 3, P in 4: R and T are adjacent (seats 2-3). So in |C|.

Arrangement 5: T in 2, P in 3, R in 4: R and T are not adjacent (seats 2 and 4 are not adjacent). So not in |C|.

Arrangement 6: T in 2, R in 3, P in 4: R and T are adjacent (seats 2-3). So in |C|.

So |C| has arrangements 1, 2, 4, 6. Wait, that's 4 arrangements. But total number of arrangements is 6. This suggests that |C| = 4, which contradicts earlier calculations. This is getting too confusing. It seems that my initial approach to calculate |A|, |B|, |C| is error-prone.

Given the time I've spent and the confusion, perhaps I should trust the gap method result for the original problem, which gave 144, and see if that makes sense.

If the gap method gives 144, and the total number of circular permutations is 720, then the number of invalid arrangements is 720 - 144 = 576. Let me see if this aligns with inclusion-exclusion.

If I assume that the number of valid arrangements is 144, then the number of invalid arrangements is 576. Let me try to calculate |A ∪ B ∪ C| = 576. Then, using inclusion-exclusion:

|A ∪ B ∪ C| = 3*|A| - 3*|A ∩ B| + |A ∩ B ∩ C| = 576.

But I need to find |A|, |A ∩ B|, etc. Let me try to calculate |A| again for the original problem.

For the original problem, n = 7, and |A| is the number of arrangements where P and R are together. Using the gap method, if we treat P and R as a block, then we have 6 entities: the block and the other 5 people. Number of circular arrangements is (6 - 1)! = 5! = 120. Multiply by 2 (for P-R or R-P) gives 240. So |A| = 240. Similarly for |B| and |C|. So |A| + |B| + |C| = 720.

Now, |A ∩ B| is the number of arrangements where P and R are together AND P and T are together, which means all three are together. As before, treating them as a block, number of circular arrangements is (5 - 1)! = 4! = 24. Multiply by 3! = 6 for internal arrangements, giving 24 * 6 = 144. So |A ∩ B| = 144. Similarly for |A ∩ C| and |B ∩ C|. So |A ∩ B| + |A ∩ C| + |B ∩ C| = 3 * 144 = 432.

Now, |A ∩ B ∩ C| is the number of arrangements where all three are together, which is the same as |A ∩ B|, which is 144. Wait, no. |A ∩ B ∩ C| is the number of arrangements where all three pairs are together, which is the same as all three being together, which is 144. So applying inclusion-exclusion:

|A ∪ B ∪ C| = 720 - 432 + 144 = 720 - 432 = 288 + 144 = 432. Wait, no: the formula is |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| = 720 - 432 + 144 = 432. Therefore, the number of valid arrangements is 720 - 432 = 288. But this contradicts the gap method result of 144. So which is correct?

Wait, let me think. If the gap method gives 144, and inclusion-exclusion gives 288, there is a conflict. Let me check with a smaller case where I can calculate manually.

Let me take n = 5, with 3 people P, R, T, and 2 others, say Q and S. Let's calculate the number of valid arrangements using both methods.

Gap method: arrange the other 2 people (Q and S) around the table. Number of circular arrangements: (2 - 1)! = 1. Number of gaps: 2. Need to place P, R, T into the gaps, no two in the same gap. But there are 2 gaps and 3 people, which is impossible. So number of valid arrangements is 0. But let's check with inclusion-exclusion.

Total number of circular permutations: (5 - 1)! = 24.

Number of invalid arrangements: |A ∪ B ∪ C|.

|A| = |B| = |C| = 2 * (5 - 2)! = 2 * 6 = 12. So 3 * 12 = 36.

|A ∩ B| = number of arrangements where all three are together: (5 - 3)! * 3! = 2! * 6 = 12. So 3 * 12 = 36.

|A ∩ B ∩ C| = 12.

So |A ∪ B ∪ C| = 36 - 36 + 12 = 12. Therefore, valid arrangements = 24 - 12 = 12. But according to the gap method, it should be 0. Contradiction again. So clearly, there is a mistake in the inclusion-exclusion approach.

This suggests that the inclusion-exclusion approach is not being applied correctly, likely due to incorrect calculation of |A|, |A ∩ B|, etc. Given the time I've spent and the confusion, I think I should trust the gap method, which gave 144 for the original problem, and see if that makes sense.

In the original problem, arranging 4 people around the table: (4 - 1)! = 6. Number of gaps: 4. Choose 3 gaps: C(4, 3) = 4. Arrange P, R, T: 3! = 6. Total: 6 * 4 * 6 = 144. This seems logical. Let me verify with n = 5, where the gap method would give: arrange 2 people, (2 - 1)! = 1. Gaps: 2. Need to place 3 people into 2 gaps: impossible. So 0, which matches the manual calculation. For n = 6, with 3 people to place, let's see: arrange 3 people around the table: (3 - 1)! = 2. Gaps: 3. Choose 3 gaps: C(3, 3) = 1. Arrange P, R, T: 3! = 6. Total: 2 * 1 * 6 = 12. Let's check with inclusion-exclusion.

Total circular permutations: (6 - 1)! = 120.

|A| = 2 * (6 - 2)! = 2 * 24 = 48. 3 * 48 = 144.

|A ∩ B| = (6 - 3)! * 3! = 6 * 6 = 36. 3 * 36 = 108.

|A ∩ B ∩ C| = 36.

So |A ∪ B ∪ C| = 144 - 108 + 36 = 72. Valid arrangements = 120 - 72 = 48. But according to the gap method, it's 12. Contradiction again. So clearly, the inclusion-exclusion approach is not being applied correctly.

Given this inconsistency, I think the gap method is more reliable here, especially since it works for the n = 4 and n = 5 cases. Therefore, I will go with the gap method result of 144 for the original problem.

But wait, let me think again. In the original problem, arranging 4 people around the table: (4 - 1)! = 6. Then, placing P, R, T in the gaps. But in circular arrangements, when you arrange the 4 people, the number of gaps is 4. Placing 3 people in 4 gaps, no two in the same gap: C(4, 3) * 3! = 4 * 6 = 24. Total arrangements: 6 * 24 = 144. This seems correct.

Alternatively, let me think of it as linear arrangements. If it were a linear arrangement, the number would be: arrange the 4 people in 4! ways, then choose 3 gaps out of 5 (since in linear, there are n + 1 gaps), but since it's circular, it's different. But the gap method for circular is to arrange the non-restricted people, then place the restricted ones in the gaps. This is a standard method, and I think it is correct.

Therefore, I will conclude that the number of valid arrangements is 144. But let me check with another approach.

Another way: total number of circular permutations: 6! = 720.

Number of ways where no two of P, R, T are adjacent: ?

First, fix P's position to remove rotational symmetry. Now, we need to arrange the remaining 6 people, but with the condition that R and T are not adjacent to P or each other.

Wait, fixing P's position, the table has 6 seats left. We need to place R and T such that they are not adjacent to P or to each other.

Wait, this might be complex, but let's try.

Fix P at a position. Now, the seats adjacent to P are two seats. We need to ensure that R and T are not in those two seats, and also not adjacent to each other.

So, first, choose seats for R and T. There are 6 seats left, excluding P's seat. The two seats adjacent to P are forbidden for R and T. So, available seats: 6 - 2 = 4 seats. Let me visualize: P is fixed, seats are labeled 1 (P), 2, 3, 4, 5, 6. Seats 2 and 6 are adjacent to P. So available seats for R and T are 3, 4, 5. Wait, no: total seats are 7, with P at seat 1. Seats 2 and 7 are adjacent to P. Wait, if it's a circular table, seat 1 is adjacent to 2 and 7. So available seats for R and T are seats 3, 4, 5, 6. Wait, 7 - 1 (P) - 2 (adjacent to P) = 4 seats. So seats 3, 4, 5, 6.

Now, we need to place R and T in these 4 seats such that they are not adjacent to each other. How many ways?

First, choose 2 seats out of 4 for R and T, such that they are not adjacent. Let's find the number of ways to choose 2 non-adjacent seats in a line of 4 seats (since seats 3, 4, 5, 6 are in a line? No, in a circular table, but since we've fixed P's position and excluded adjacent seats, the remaining seats are in a line? Wait, no. If P is fixed at seat 1, seats 2 and 7 are adjacent to P. The remaining seats are 3, 4, 5, 6. These are in a line? No, in a circular table, seat 3 is adjacent to 4 and 2, seat 4 is adjacent to 3 and 5, seat 5 is adjacent to 4 and 6, seat 6 is adjacent to 5 and 7. But seat 7 is adjacent to P (seat 1) and seat 6. But we have excluded seat 2 and 7 for R and T. Wait, no, we have excluded seats 2 and 7 (adjacent to P) for R and T. So R and T can be placed in seats 3, 4, 5, 6. Now, in this set of seats, seat 3 is adjacent to 4, seat 4 to 5, seat 5 to 6, and seat 6 is adjacent to 5 and 7 (but 7 is excluded). So in the available seats 3, 4, 5, 6, the adjacency is between 3-4, 4-5, 5-6. So it's like a line of 4 seats: 3-4-5-6. So the number of ways to choose 2 non-adjacent seats in a line of 4 seats is C(4, 2) - 3 = 6 - 3 = 3. Wait, in a line of 4 seats, the number of adjacent pairs is 3 (3-4, 4-5, 5-6). Total pairs: 6. Non-adjacent pairs: 6 - 3 = 3. These are (3,5), (3,6), (4,6). So 3 ways.

For each such choice, we can arrange R and T in 2! ways. So total ways: 3 * 2 = 6.

Now, after placing R and T, the remaining 4 people (Q, S, T, U? Wait, original problem has 7 people: P, R, T, and 4 others. Wait, in the original problem, there are 7 people: P, R, T, and 4 others. So after fixing P, we have 6 seats left, and we placed R and T in 2 of them, leaving 4 seats for the other 4 people. Wait, no: total people are 7. Fix P, then we have 6 people left: R, T, and 4 others. We placed R and T in 2 seats, so the remaining 4 seats are for the other 4 people. The number of ways to arrange them is 4!.

So total number of arrangements: 3 * 2 * 4! = 6 * 24 = 144. Which matches the gap method result! So this confirms that the number of valid arrangements is 144.

Therefore, the answer is 144. But wait, the question says "Seating arrangements which are rotations of each other are treated as the same." So we have already accounted for that by fixing P's position, or by using the gap method which inherently accounts for circular permutations. Therefore, the final answer is 144.

But let me check with the inclusion-exclusion result. Earlier, using inclusion-exclusion, I got 288, but that must have been incorrect. Now, with this method, it's 144. Since this method is verified with a smaller case, I trust it.

Therefore, the number of ways is 144, so the answer is \boxed{144}.

**Final Answer**
\boxed{144}

To determine the number of ways 7 people can sit around a round table such that **Pierre (P), Rosa (R), and Thomas (T)** are **not adjacent to each other**, we use a **gap method** that accounts for the circular nature of the table and the restriction on adjacency.

---

### Step 1: Total Circular Permutations

In a circular arrangement of $ n $ people, the number of distinct seating arrangements is:

$$
(n - 1)! = 6! = 720
$$

This is because rotations of the same arrangement are considered identical.

---

### Step 2: Apply the Gap Method

We want to **avoid** any two of P, R, and T sitting next to each other. A standard technique for such problems is to:

1. **Fix the positions of the other people** (those not in the restricted group).
2. **Place the restricted people (P, R, T)** in the **gaps** between the others, ensuring that no two of them are adjacent.

#### Step 2.1: Arrange the Other 4 People

We first arrange the 4 people who are **not** P, R, or T. Since the table is circular, the number of distinct arrangements is:

$$
(4 - 1)! = 3! = 6
$$

These 4 people create **4 gaps** between them (since it's a circle).

#### Step 2.2: Choose Gaps for P, R, T

We need to place P, R, and T into **3 of these 4 gaps**, ensuring that **no two of them are in the same gap** (to prevent adjacency). The number of ways to choose 3 gaps from 4 is:

$$
\binom{4}{3} = 4
$$

#### Step 2.3: Arrange P, R, T in the Chosen Gaps

Once the gaps are chosen, we can arrange P, R, and T in those 3 gaps in:

$$
3! = 6 \text{ ways}
$$

---

### Step 3: Total Valid Arrangements

Now, multiply all the components:

- Ways to arrange the 4 non-restricted people: $ 3! = 6 $
- Ways to choose 3 gaps from 4: $ \binom{4}{3} = 4 $
- Ways to arrange P, R, T in the chosen gaps: $ 3! = 6 $

$$
\text{Total valid arrangements} = 6 \times 4 \times 6 = 144
$$

---

### Final Answer

$$
\boxed{144}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways 7 people can sit around a round table, but with the condition that Pierre, Rosa, and Thomas (let me call them P, R, T for short) can't sit next to each other. And rotations are considered the same arrangement. Hmm, circular permutations can be tricky because rotations are considered identical, so usually we fix one person's position to avoid counting rotations multiple times. Let me recall the formula for circular permutations: for n people, it's (n-1)! because fixing one person and arranging the rest. But here, there are restrictions.

First, maybe I should find the total number of seating arrangements without any restrictions and then subtract the ones where at least two of P, R, T are sitting together. Wait, but inclusion-exclusion might be needed here because subtracting the cases where two are together might overcount or undercount. Alternatively, maybe it's better to use the principle of inclusion-exclusion. Let me think.

Alternatively, maybe I can approach it by first arranging the other people and then placing P, R, T in the gaps. That might be a better approach. Let me try that.

In circular permutations, arranging people around a table, if we want to ensure that certain people are not adjacent, sometimes we can use the gap method. Let me recall: if we have n people, the number of ways to arrange them around a table is (n-1)! If we want to place some people so that they are not adjacent, we can first arrange the others and then place the restricted people in the gaps.

So here, there are 7 people total. Let me consider the 3 people P, R, T who can't sit together. So maybe first arrange the other 4 people around the table, then place P, R, T in the gaps between them such that no two are adjacent.

But wait, in circular arrangements, arranging the 4 people first. Let me check: if we fix one person's position to account for rotational symmetry, then arrange the rest. So arranging 4 people around a table: (4-1)! = 6 ways. Then, between these 4 people, there are 4 gaps (since it's a circle). Wait, if there are 4 people around a table, there are 4 gaps between them. So if we want to place P, R, T into these gaps, each in separate gaps, so that no two are adjacent. Since there are 4 gaps and 3 people to place, the number of ways would be C(4,3) * 3! (choosing 3 gaps out of 4 and arranging the 3 people in them). So total arrangements would be 6 * C(4,3) * 3! = 6 * 4 * 6 = 144? Wait, but let me check again.

Wait, first arranging the 4 people: (4-1)! = 6. Then, the number of ways to place P, R, T into the gaps: since there are 4 gaps, and we need to choose 3 gaps, which is C(4,3) = 4, and then arrange P, R, T in those gaps, which is 3! = 6. So total is 6 * 4 * 6 = 144. But wait, the total number of circular permutations without restrictions is (7-1)! = 720. So 144 is less than 720, but is this the correct answer? Wait, but maybe I made a mistake here.

Wait, hold on. Let me think again. When arranging the 4 people around the table, fixing one person's position, so that rotations are accounted for. Then, the number of ways to arrange the other 3 people is 3! = 6? Wait, no. Wait, if we fix one person's position, then the remaining 3 people can be arranged in 3! ways. Wait, but the total number of arrangements for 4 people around a table is (4-1)! = 6. So that's correct. So arranging the 4 people: 6 ways. Then, inserting P, R, T into the gaps. Since there are 4 gaps between the 4 people, and we need to place 3 people into different gaps. So the number of ways is C(4,3) * 3! = 4 * 6 = 24. Then total arrangements would be 6 * 24 = 144. So that would be the answer? But let me check if this is correct.

Alternatively, maybe I need to consider that when arranging the 4 people, the number of gaps is 4, and inserting P, R, T into those gaps. But since the table is circular, inserting into gaps is okay. But let me think of another approach to verify.

Alternatively, total number of seating arrangements without restriction is 6! / 7? Wait no, circular permutations for n people is (n-1)! So for 7 people, it's 6! = 720. So if I use inclusion-exclusion, maybe I can compute the total number of arrangements where at least two of P, R, T are sitting together, and subtract that from 720.

But inclusion-exclusion can get complicated here. Let me try that approach as well to check.

First, total number of arrangements: 6! = 720.

Now, let me compute the number of arrangements where at least one pair of P, R, T are sitting together. Let me denote:

Let A be the set of arrangements where P and R are sitting together.

B be the set where P and T are sitting together.

C be the set where R and T are sitting together.

We need to compute |A ∪ B ∪ C|, and subtract that from total to get the desired number.

By inclusion-exclusion:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.

So first compute |A|, |B|, |C|. Since the problem is symmetric for P, R, T, each of these will be equal. So |A| = |B| = |C|.

Compute |A|: number of arrangements where P and R are sitting together. For circular tables, treating P and R as a single entity. But since it's circular, the number of arrangements is (n-1)! / n? Wait, no. Wait, when we have two people sitting together, we can consider them as a single "block". So for circular arrangements, the number of ways to arrange n-1 objects (the block and the other n-2 people) is (n-2)! But since the block can be arranged in 2 ways (P-R or R-P). Wait, but in circular permutations, when we have a block, how does that work?

Wait, let me recall. For linear arrangements, if we have two people together, we treat them as a block, so total n-1 elements, which can be arranged in (n-1)! ways, multiplied by 2 for the internal arrangement. But for circular arrangements, fixing one position to account for rotation. Let me think.

Alternatively, the formula for circular permutations with a block: if we have a block of k people, then the number of circular arrangements is (n - k)! * k! ? Wait, no. Wait, if we have a block, then effectively, we have (n - k + 1) objects to arrange? Wait, maybe not. Let me think again.

Suppose we have n people around a table. If we consider two people as a single block, then we have n - 1 objects to arrange (the block and the other n - 2 people). However, in circular permutations, the number of arrangements is (number of objects - 1)! So (n - 2)! And then multiply by 2 for the internal arrangement of the block. Therefore, |A| = 2 * (n - 2)! So for n = 7, |A| = 2 * 5! = 2 * 120 = 240. Wait, but hold on, total number of circular permutations is 6! = 720. If |A| is 240, then |A| + |B| + |C| = 3 * 240 = 720. But then when we subtract the intersections, etc., but let me check if this is correct.

Wait, but if we have 3 sets A, B, C each with 240 elements, then |A ∪ B ∪ C| would be 3*240 - ... but that seems like it might be overcounting. However, let me check with n=7.

Wait, but let me think again. Maybe my formula for |A| is wrong. Let me think of another way. For circular permutations, the number of ways where P and R are sitting together: fix P's position to remove rotational symmetry. Then R can be either to the left or right of P, so 2 choices. Then the remaining 5 people can be arranged in 5! ways. Wait, but if we fix P's position, then total arrangements are 6! / 7? Wait, no. Wait, fixing P's position, the total number of arrangements is 6! (since the rest can be arranged in 6! ways). Wait, but if we fix P's position, then the number of arrangements where R is adjacent to P is 2 * 5! (since R can be on the left or right, and the rest can be arranged in 5! ways). Therefore, the number of arrangements where P and R are adjacent is 2 * 5! = 240. But total number of arrangements is 6! = 720. So that seems correct. Therefore, |A| = |B| = |C| = 240. Then moving on.

Now, compute |A ∩ B|: arrangements where both P and R are together and P and T are together. Which implies that P, R, T are all sitting together. Because if P is adjacent to both R and T, then R and T are on either side of P, so all three are together. So the intersection of A and B is the set of arrangements where P, R, T are all sitting together. Similarly for other intersections.

So |A ∩ B| = |A ∩ C| = |B ∩ C| = number of arrangements where all three P, R, T are sitting together. Let me compute that.

Again, treating P, R, T as a single block. Then, we have 7 - 3 + 1 = 5 objects to arrange: the block and the other 4 people. For circular arrangements, the number of ways is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways (since P, R, T can be in any order within the block). Therefore, total is 24 * 6 = 144. Wait, but let me check with fixing positions again.

Alternatively, fix P's position. Then, since P, R, T are all together, we can think of the block as P, R, T in some order. If we fix P's position, then the block can be arranged in 2 ways: either R and T are on the left and right of P, or T and R are on the left and right. Wait, no. Wait, if we fix P's position, then the block can be arranged in 2! ways? Wait, no. Wait, if we fix P's position, then the block can be arranged in 2! ways? Wait, no. Wait, if the block is P, R, T, then once P is fixed, the other two can be arranged in 2! ways. But also, the block can be placed in two different directions? Wait, no. Wait, if we fix P's position, then the block can extend clockwise or counter-clockwise? Wait, no. Since it's a circular table, fixing P's position, the block can be arranged in 2! ways for the other two people. Wait, but actually, the entire block can be arranged in 3! ways, but since we fixed P's position, maybe?

Wait, this is getting confusing. Let me use the standard formula. If we have a block of 3 people, then the number of circular arrangements is (n - k)! * k! where n is total people, k is the block size? Wait, no. Wait, earlier, for a single block, we had (n - 1)! * 2 for two people. Wait, maybe not. Let me think again.

If we have a block of size k, then the number of circular arrangements is (n - k)! * k! ?

Wait, no. Wait, for linear arrangements, it's (n - k + 1)! * k! But for circular arrangements, it's different. Let me think of fixing one position. Suppose we fix one person's position to account for circular symmetry. Let's fix P's position. Then, the block of P, R, T can be arranged in 2! ways (since P is fixed, R and T can be arranged in 2! ways). But also, the block can be placed in different positions? Wait, no. If we fix P's position, then the block is fixed in terms of position? Wait, no. Wait, if we fix P's position, then the block can be arranged in two directions? Wait, maybe not. Let me think.

Alternatively, if we consider the block as a single entity, then with P fixed, the block can be placed in such a way that P is fixed, and the rest of the block is arranged. Wait, this is getting too tangled. Let me use the formula for circular permutations with a block. If we have a block of size k, then the number of circular arrangements is (n - k)! * k! ?

Wait, no. Wait, for example, if n = 3, and k = 3, then the number of circular arrangements is 1 (since all are together), but according to the formula (n - k)! * k! = 0! * 3! = 6, which is wrong. So that formula is incorrect.

Alternative approach: when considering circular permutations with a block, we can treat the block as a single entity. So, if we have n people, and we have a block of size k, then the number of circular arrangements is (n - k)! * k! / n? Wait, no. Wait, in linear arrangements, it's (n - k + 1)! * k! But for circular arrangements, it's different. Let me think of fixing one position to remove rotational symmetry.

Suppose we fix one person's position. Let's fix P's position. Then, the block of P, R, T can be arranged in 2! ways (since P is fixed, R and T can be arranged in 2! ways). Then, the remaining 4 people can be arranged in 4! ways. So total arrangements would be 2! * 4! = 2 * 24 = 48. But wait, but if we fix P's position, then the total number of arrangements where P, R, T are all together is 2! * 4! = 48. However, the total number of circular arrangements is 6! = 720. Wait, but if we fix P's position, the total number of arrangements is 6! / 7? No, fixing P's position, the total number of arrangements is 6! / 7? Wait, no. Wait, fixing one person's position in circular permutations, the number of arrangements is (n - 1)! So for n = 7, fixing P's position, the total number of arrangements is 6! = 720. Wait, no! Wait, no. Wait, if you fix one person's position, the number of arrangements is (n - 1)! So for n = 7, fixing P's position, the number of arrangements is 6! = 720? Wait, no! Wait, no. Wait, if you fix one person's position, the number of arrangements is (n - 1)! So for n = 7, fixing one person, the number of arrangements is 6! = 720? Wait, no! Wait, no. Wait, (n - 1)! is the total number of circular permutations. If you fix one person's position, then the rest can be arranged in (n - 1)! / n? No, no. Wait, confusion here.

Let me recall: the number of circular permutations of n people is (n - 1)! because rotations are considered the same. If we fix one person's position, then the number of arrangements is (n - 1)! / n * n? Wait, no. Wait, fixing one person's position effectively removes rotational symmetry, so the number of arrangements is (n - 1)! So for example, for n = 3, fixing one person, the other two can be arranged in 2! ways, which is (3 - 1)! = 2. Correct. So for n = 7, fixing one person's position, the number of arrangements is 6! = 720? Wait, no! Wait, (n - 1)! is 6! = 720. But if you fix one person's position, then the number of arrangements is (n - 1)! So for n = 7, fixing one person, the number of arrangements is 6! = 720? Wait, that can't be. Wait, no. Wait, if you fix one person's position, then the number of arrangements is (n - 1)! So for n = 7, it's 6! = 720? Wait, but that would mean that the total number of circular permutations is 720, which is correct. So if we fix one person's position, the number of arrangements is 6! = 720. Wait, but that seems contradictory. Wait, no. Wait, the total number of circular permutations is (n - 1)! So for n = 7, it's 6! = 720. If we fix one person's position, then the number of arrangements is (n - 1)! So that is the same as the total number of circular permutations. Wait, that seems redundant. Wait, no. Wait, fixing one person's position is a way to count circular permutations. So if you fix one person's position, then the rest can be arranged in (n - 1)! ways. Wait, no. Wait, if you fix one person's position, then the number of ways to arrange the rest is (n - 1)! / n? No. Wait, confusion arises here. Let me check with n = 3. If you fix one person's position, then the other two can be arranged in 2! ways, which is (3 - 1)! = 2. So yes, fixing one person's position, the number of arrangements is (n - 1)! So for n = 7, fixing one person's position, the number of arrangements is 6! = 720. Wait, but that's the same as the total number of circular permutations. Wait, but that's because fixing one person's position is a way to count them. So if you fix one person's position, you are effectively counting the circular permutations as linear permutations of the remaining people. So, okay, that seems correct.

So, going back, if we fix P's position, then the number of arrangements where P, R, T are all together is calculated as follows: since P is fixed, the block of P, R, T can be arranged in 2! ways (since P is fixed, R and T can be on either side). Wait, but actually, the block can be arranged in 2! ways? Wait, if P is fixed, then the block can be P-R-T or P-T-R? Wait, but since it's a circular table, once P is fixed, the block can be arranged in two different ways: R to the left of P and T to the right, or T to the left and R to the right. So that's 2 ways. Then, the remaining 4 people can be arranged in 4! ways. Therefore, total arrangements where P, R, T are all together is 2 * 4! = 2 * 24 = 48. Therefore, the number of circular arrangements where P, R, T are all together is 48. But wait, but earlier I thought it was 144. Which is correct?

Wait, if we don't fix positions, then the number of circular arrangements where P, R, T are all together is calculated by treating them as a block. So, number of objects is 5 (the block + 4 others). Number of circular arrangements is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways. So total is 24 * 6 = 144. But if we fix P's position, we get 48. Which is different. So which is correct?

Wait, if we fix P's position, then the total number of arrangements is 6! = 720. But if we don't fix positions, the total number of circular arrangements is 6! = 720. So if we fix P's position, the number of arrangements where P, R, T are all together is 48. Therefore, the total number without fixing positions would be 48 * 7 / 1? Wait, no. Wait, if we fix P's position, we are considering only 1/n of the total arrangements? Wait, no. Wait, fixing P's position is a way to count all circular arrangements. So if we fix P's position, the number of arrangements where P, R, T are all together is 48, which would be the same as the total number of circular arrangements where P, R, T are all together. Wait, but according to the other method, treating the block as a single entity, we get 144. Which is different. So which is correct?

Wait, let me think. If we don't fix positions, the number of circular arrangements where P, R, T are all together is calculated as follows: consider the block of P, R, T as a single entity. Then, we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways. So total is 24 * 6 = 144. However, if we fix P's position, we get 48. But 144 is 3 times 48. Why?

Ah! Because when we fix P's position, we are considering only arrangements where P is fixed. However, in the total number of circular arrangements where P, R, T are all together, P can be in any position. So if we fix P's position, we are counting only the arrangements where P is fixed, but in reality, there are multiple positions where P can be. Wait, but in circular permutations, fixing P's position accounts for all rotations. Therefore, if we fix P's position, the number of arrangements where P, R, T are all together is 48, which should be equal to the total number of such arrangements divided by 7 (since there are 7 positions for P). Wait, but if total number is 144, then 144 / 7 is not an integer. Which is a problem. Therefore, one of the methods is wrong.

Wait, this suggests that there is a mistake in one of the approaches. Let me check again.

First approach: treating the block as a single entity. Number of circular arrangements is (5 - 1)! * 3! = 24 * 6 = 144. But if we fix P's position, we get 48. Let me check with n=3. If n=3, and we want the number of circular arrangements where all 3 are together. Well, there's only 1 arrangement, since they are all together. But according to the first method: treating the block as a single entity, number of circular arrangements is (1 - 1)! * 3! = 0! * 6 = 1 * 6 = 6. Which is wrong. So clearly, that formula is incorrect. Therefore, my previous approach was wrong.

So, the formula for circular permutations with a block is not (n - k)! * k! So that approach is invalid. Therefore, need to think differently.

Alternative approach: For circular arrangements, the number of ways where a specific set of people are all together can be calculated by considering the block as a single entity. However, in circular permutations, when you have a block, the formula is different. Let me check.

In linear arrangements, the number of ways to arrange n people with a block of k people together is (n - k + 1)! * k!.

But for circular arrangements, it's a bit different. Let me think. If we have a block of k people, then the number of circular arrangements is (n - k)! * k! / n? Wait, not sure. Wait, here's a method: fix the block's position to account for circular symmetry. For example, fix the block in a specific position, then arrange the rest. But since the table is circular, fixing the block's position removes rotational symmetry.

Wait, suppose we fix the block of P, R, T in a specific position. Then, the number of ways to arrange the block is 3! (since they can be in any order). Then, the remaining 4 people can be arranged in 4! ways. However, since we fixed the block's position, we don't have to worry about rotations. Therefore, total number of arrangements is 3! * 4! = 6 * 24 = 144. But this seems to contradict the previous result when fixing P's position. Wait, but if we fix the block's position, then we are not accounting for rotations. However, in circular permutations, fixing the block's position is equivalent to considering rotational symmetry. Therefore, the total number of circular arrangements where P, R, T are all together is 3! * 4! = 144. However, when we fixed P's position earlier, we got 48. Which is different. So why?

Because when we fixed P's position, we were considering only arrangements where P is fixed, but in reality, the block can be anywhere. However, if we fix P's position, then the block can be arranged in two different ways (as discussed earlier), and the rest can be arranged in 4! ways. So 2 * 4! = 48. But according to the other method, it's 144. Which is 3 times larger. Where is the discrepancy?

Ah! Because when we fix P's position, we are considering only arrangements where P is fixed, but in reality, the block can be placed in different positions relative to P. Wait, no. If we fix P's position, then the block must include P. So the block is fixed with P in a specific position, and the rest of the block (R and T) can be arranged in 2 ways. Then, the rest of the people can be arranged in 4! ways. Therefore, total is 2 * 4! = 48. However, if we don't fix P's position, the total number of arrangements where P, R, T are all together is 144. Therefore, 144 is equal to 7 * 48 / something? Wait, no. Wait, if we fix P's position, we are considering 1/7 of all possible arrangements. Therefore, if the total number of arrangements where P, R, T are all together is 144, then fixing P's position would give 144 / 7? But 144 / 7 is not an integer. Which is a problem. Therefore, there must be an error in one of the methods.

Wait, let me think again. Let me use the formula for circular permutations with a block. Suppose we have n people, and we want to count the number of circular arrangements where a specific set of k people are all together. The formula is (n - k)! * k! / n? Wait, no. Wait, here's a resource I remember: when arranging people around a circular table with a group of people together, the number of arrangements is (n - k)! * k! / n? Wait, not sure. Wait, let me think of it as follows.

In circular permutations, the number of ways to arrange n people is (n - 1)! If we want to have a specific group of k people together, we can consider them as a single entity. Then, we have (n - k + 1) entities to arrange. However, in circular permutations, the number of ways to arrange (n - k + 1) entities is (n - k)! So total number of arrangements is (n - k)! * k! (for arranging the group internally). But wait, for example, with n = 3, k = 3: (3 - 3)! * 3! = 1 * 6 = 6, which is wrong because there's only 1 circular arrangement. So that formula is wrong.

Wait, maybe the formula is (n - k)! * k! / n? For n = 3, k = 3: (0)! * 6 / 3 = 1 * 6 / 3 = 2, which is still wrong. Hmm.

Alternative approach: Let me think of the number of ways where P, R, T are all together. First, consider them as a single block. Then, we have 5 entities: the block and the other 4 people. The number of circular arrangements of these 5 entities is (5 - 1)! = 4! = 24. Then, the block can be arranged in 3! ways. So total is 24 * 6 = 144. However, in this calculation, are we overcounting? For example, if the block is considered as a single entity, but in circular arrangements, rotating the entire block doesn't create a new arrangement. Wait, but since we already accounted for circular arrangements by using (5 - 1)! for the entities, then multiplying by the internal arrangements of the block should be correct. However, when n = 3, this would give (3 - 3)! * 3! = 1 * 6 = 6, which is wrong. But for n = 3, the correct number is 1. So why is there a discrepancy?

Ah! Because when the block is the entire set of people, the formula (n - k)! * k! counts linear arrangements, not circular. Wait, no. Wait, if we have n = 3, k = 3, then treating them as a single block, we have 1 entity. The number of circular arrangements of 1 entity is 1 (since there's only one way to arrange a single entity around a table). Then, multiply by k! = 6, which would give 6, which is wrong. Therefore, clearly, this approach is incorrect.

Therefore, there must be a different way. Let me think again. Maybe the formula for circular permutations with a block is different. Let me refer to standard results.

Upon recalling, the number of ways to arrange n people around a circular table with a specific group of k people sitting together is (n - k)! * k! / n? Wait, no. Wait, here's a better way. Let me fix the position of the block to eliminate rotational symmetry. Suppose we fix the block in a specific position. Then, the number of ways to arrange the block is k! and the rest can be arranged in (n - k)! ways. However, since we fixed the block's position, we don't need to divide by anything. Therefore, total number of arrangements is k! * (n - k)! But for n = 3, k = 3, this would be 6 * 1 = 6, which is still wrong. But the correct number is 1. So this is not correct.

Wait, but if we fix the block's position, then the number of circular arrangements is (n - k)! * k! / n? No. Wait, confusion reigns. Let me try with n = 3, k = 2. Suppose we want the number of circular arrangements where two specific people are together. Let's say P and R. Fix P's position. Then R can be to the left or right of P, so 2 choices. The third person, T, can be arranged in 1! way. So total is 2 * 1 = 2. But the total number of circular arrangements is 2! = 2. Wait, but if we don't fix positions, the number of circular arrangements where P and R are together is 2. Which is correct. However, according to the formula (n - k)! * k! = (3 - 2)! * 2! = 1 * 2 = 2, which matches. Wait, but earlier with n = 3, k = 3, it would be (3 - 3)! * 3! = 1 * 6 = 6, which is wrong. But if we fix the block's position, then for n = 3, k = 3, fixing the block's position, we have 1 way to arrange the block (since it's the entire table), and the rest is 0! = 1, so total is 1 * 6 = 6? But that's wrong. Wait, but if we fix the block's position, then the number of arrangements is k! * (n - k)! = 6 * 1 = 6, but in reality, there's only 1 circular arrangement. So why is there a discrepancy?

Ah! Because when we fix the block's position, we are not accounting for rotational symmetry. Wait, if we fix the block's position, then we are effectively making it a linear arrangement. Wait, no. If we fix the block's position on the table, then the rest of the people are arranged relative to it. However, in circular permutations, fixing the block's position removes rotational symmetry. Therefore, the number of arrangements should be k! * (n - k)! But for n = 3, k = 3, this would be 6 * 1 = 6, but there's only 1 circular arrangement. Therefore, this formula is wrong.

Wait, but if we fix the block's position, then the number of arrangements is k! * (n - k)! / something? Wait, no. Wait, if we fix the block's position, then the number of arrangements is k! * (n - k)! / n? No. Wait, confusion. Let me think of n = 3, k = 2. If we fix the block of P and R, then the third person T can be arranged in 1! ways. But since we fixed the block's position, the total number of arrangements is 2! * 1! = 2, which matches the correct number. However, if we don't fix the block's position, the number of circular arrangements where P and R are together is 2, which is equal to (n - k)! * k! = 1! * 2! = 2. So that works. But for n = 3, k = 3, it would be (n - k)! * k! = 0! * 3! = 6, which is wrong. But if we fix the block's position, we get 3! * 0! = 6, which is wrong. So why does it work for k = 2 but not for k = 3?

Ah! Because when k = n, the formula (n - k)! * k! = 0! * n! = n! which is the total number of circular permutations, but in reality, there is only 1 circular permutation. So clearly, the formula is not correct for k = n. Therefore, there must be a different formula.

Wait, but when k = n, the number of circular arrangements where all people are together is 1 (since they are all together). But according to the formula (n - k)! * k! = n! which is wrong. Therefore, the formula is only valid when k < n? Or maybe there's a different approach.

Wait, maybe the formula is (n - k)! * k! / n? For n = 3, k = 3: 0! * 6 / 3 = 2, which is still wrong. For n = 3, k = 2: 1! * 2 / 3 = 2/3, which is wrong. So that doesn't work.

This is getting too confusing. Maybe I should go back to the original problem and try to use the first method where I arranged the other people first and then placed P, R, T in the gaps. Earlier, I calculated that as 6 * 4 * 6 = 144. But if the total number of circular permutations is 720, then 144 is 1/5 of that. But let me check with another approach.

Alternatively, let me think of the total number of ways where no two of P, R, T are adjacent. This is equal to total number of circular permutations minus the number of permutations where at least two are adjacent. But earlier, I tried inclusion-exclusion but got stuck on calculating |A ∩ B|.

Alternatively, let me try to use the gap method again. Let me try to calculate it properly.

First, arrange the other 4 people around the table. Since it's a circular table, the number of ways is (4 - 1)! = 6. Now, between these 4 people, there are 4 gaps. We need to place P, R, T into these gaps such that no two are in the same gap (to ensure they are not adjacent). The number of ways to choose 3 gaps out of 4 is C(4, 3) = 4. Then, arrange P, R, T in these gaps: 3! = 6. So total number of ways is 6 * 4 * 6 = 144. So this gives 144. But earlier, when I tried inclusion-exclusion, I was confused about the |A ∩ B| term. Let me try to proceed with inclusion-exclusion now, using this result.

If the answer is 144, then the total number of valid arrangements is 144. But let me check with inclusion-exclusion.

Total number of circular permutations: 6! = 720.

Number of arrangements where at least one pair of P, R, T are adjacent: |A ∪ B ∪ C|.

As before, |A| = |B| = |C| = 2 * 5! = 240. Wait, earlier I thought that fixing P's position gives 2 * 5! = 240, but that seems high. Wait, if total number of circular permutations is 720, and |A| is 240, then |A| + |B| + |C| = 720. Then, subtracting intersections, etc. But if |A ∩ B| is the number of arrangements where P and R are together AND P and T are together, which implies all three are together. So |A ∩ B| = number of arrangements where all three are together. Let me calculate that.

If all three are together, then as per the gap method, if we consider them as a block, then number of circular arrangements is (5 - 1)! * 3! = 24 * 6 = 144? Wait, but earlier we saw that this might be wrong. But if we use the gap method, we have 144 as the number of valid arrangements, which would mean that the number of invalid arrangements is 720 - 144 = 576. But according to inclusion-exclusion, if |A ∪ B ∪ C| = 3*240 - 3*|A ∩ B| + |A ∩ B ∩ C|. Let me assume that |A ∩ B| = number of arrangements where all three are together, which we need to calculate.

If we assume that |A ∩ B| = number of arrangements where all three are together, which is 144, then |A ∩ B| = |A ∩ C| = |B ∩ C| = 144. Then, |A ∩ B ∩ C| is also the number of arrangements where all three are together, which is 144. Wait, but that can't be. Because |A ∩ B ∩ C| is the set of arrangements where all three pairs are together, which is the same as all three being together. So |A ∩ B ∩ C| = 144.

Therefore, |A ∪ B ∪ C| = 3*240 - 3*144 + 144 = 720 - 432 + 144 = 720 - 432 = 288 + 144 = 432? Wait, 3*240 = 720, then subtract 3*144 = 432, so 720 - 432 = 288, then add back 144, so 288 + 144 = 432. Therefore, |A ∪ B ∪ C| = 432. Therefore, the number of valid arrangements is total - invalid = 720 - 432 = 288. But this contradicts the gap method result of 144. So which is correct?

This suggests that there is a mistake in one of the methods. Let me check with smaller numbers. Let me take n = 4, with 3 people who can't sit together. Wait, maybe that's too small. Let me try n = 4, with 3 people P, R, T and one other person, say Q. How many circular arrangements where P, R, T are not all adjacent? Wait, but in n = 4, if we have 3 people who can't sit together... Wait, in a circular table of 4 people, if we want to ensure that P, R, T are not all adjacent. Wait, but in a table of 4, any three people will be adjacent if they are seated together. Wait, no. In a circular table of 4, if three people are seated, they are adjacent if they are next to each other. But with 4 people, the number of ways to seat them is (4-1)! = 6. How many of these have P, R, T all together? Let's calculate.

Using the gap method: arrange the other person Q around the table. Since there's only one person, arranging Q around the table: (1-1)! = 1 way. Then, there are 1 gap between Q. Wait, but we need to place P, R, T into gaps. But there's only 1 gap, and we need to place 3 people. Which is impossible. Therefore, the number of valid arrangements is 0? But that can't be. Wait, in a circular table of 4 people, if we have 3 people who cannot sit together, but since there are only 4 seats, and 3 people, they must be seated with one seat between them? Wait, no. Wait, if you have 4 seats, and 3 people, then one seat is empty. The three people can be seated such that no two are adjacent? Wait, in a circular table of 4 seats, can 3 people be seated with no two adjacent? No, because if you have 3 people, each person has two neighbors. In a circle of 4, if you seat 3 people, at least two of them must be adjacent. Because with 3 people, there are 3 seats occupied and 1 empty. The empty seat is between two people, so those two people are adjacent. Therefore, it's impossible to seat 3 people around a table of 4 without having at least two adjacent. Therefore, the number of valid arrangements is 0. Which matches the gap method result. But according to inclusion-exclusion, let's see:

Total number of circular permutations: 6.

Number of arrangements where at least one pair of P, R, T are adjacent. Let's compute |A ∪ B ∪ C|.

Each |A|, |B|, |C|: number of arrangements where a specific pair is together. For n = 4, fixing one person's position, say Q. Then, the number of arrangements where P and R are together: treat P and R as a block. Then, we have Q and the block. Number of circular arrangements: (2 - 1)! = 1! = 1. Multiply by 2 (for P-R or R-P): 2. Then, the total number of arrangements where P and R are together is 2. Similarly for |B| and |C|. So |A| + |B| + |C| = 3 * 2 = 6. Now, |A ∩ B|: arrangements where P and R are together AND P and T are together, which means all three are together. Number of such arrangements: treat P, R, T as a block. Then, we have Q and the block. Number of circular arrangements: (2 - 1)! = 1. Multiply by 3! = 6 for internal arrangements. So total is 6. Therefore, |A ∩ B| = |A ∩ C| = |B ∩ C| = 6. Then, |A ∩ B ∩ C| is also 6. Therefore, |A ∪ B ∪ C| = 6 - 3*6 + 6 = 6 - 18 + 6 = -6, which is impossible. Clearly, something is wrong here.

This suggests that my inclusion-exclusion approach is flawed. Therefore, there must be a mistake in how I'm calculating |A|, |B|, |C|, etc. For n = 4, the total number of circular permutations is 6. If I calculate |A| as 2, then 3*2 = 6, which equals the total number of permutations. Therefore, |A ∪ B ∪ C| = 6, so the number of valid arrangements is 6 - 6 = 0, which is correct. But according to the inclusion-exclusion formula, I had |A ∪ B ∪ C| = 3*2 - 3*6 + 6 = -6, which is wrong. Therefore, my calculation of |A ∩ B| was incorrect.

So, what is the correct value of |A ∩ B|? For n = 4, |A ∩ B| is the number of arrangements where both P and R are together and P and T are together, which means all three are together. How many such arrangements are there? If we treat P, R, T as a block, then we have the block and Q. Number of circular arrangements: (2 - 1)! = 1. Multiply by the number of internal arrangements of the block: 3! = 6. So total is 6. But the total number of circular permutations is 6, so this suggests that all arrangements have all three together, which is not true. Wait, but in reality, when n = 4, any arrangement of P, R, T, Q must have at least two of P, R, T together. Wait, but if you have 3 people and 1 empty seat, then in any arrangement, at least two of the three people are adjacent. Therefore, all arrangements have at least one pair together, so |A ∪ B ∪ C| = 6, and the number of valid arrangements is 0. But according to the inclusion-exclusion formula, if |A| = 2, |B| = 2, |C| = 2, |A ∩ B| = 6, |A ∩ C| = 6, |B ∩ C| = 6, and |A ∩ B ∩ C| = 6, then:

|A ∪ B ∪ C| = 3*2 - 3*6 + 6 = 6 - 18 + 6 = -6, which is impossible. Therefore, clearly, my calculation of |A ∩ B| is wrong.

Wait, what is the correct value of |A ∩ B|? For n = 4, |A ∩ B| is the number of arrangements where both P and R are together and P and T are together. Which means all three are together. How many such arrangements are there? Let's list them. Fix Q's position. Then, the block of P, R, T can be arranged in 2 ways: clockwise or counter-clockwise? Wait, no. If Q is fixed, then the block can be arranged in 2 ways: P-R-T or P-T-R, but since it's a circular table, fixing Q's position, the block can be placed in two different ways? Wait, no. If Q is fixed, then the block of P, R, T occupies the other three seats. Since it's a circular table, the block can be arranged in 2 ways: clockwise or counter-clockwise? Wait, no. If Q is fixed, then the three seats adjacent to Q are in a line? No, in a circular table with Q fixed, the other three seats are arranged around Q. If we have a block of P, R, T, then they can be arranged in 3! = 6 ways. But since the table is circular, but Q is fixed, so rotations are already accounted for. Therefore, the number of arrangements where P, R, T are all together is 3! = 6. But the total number of circular permutations is 6, so this suggests that all arrangements have P, R, T together, which is true, because with 3 people and 1 empty seat, they must be adjacent. Therefore, |A ∩ B| = 6. But then, in inclusion-exclusion, we have:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| = 2 + 2 + 2 - 6 - 6 - 6 + 6 = 6 - 18 + 6 = -6, which is impossible. Therefore, clearly, my calculation of |A| is wrong.

Wait, earlier I said |A| = 2 for n = 4. Let me recalculate |A|. For n = 4, |A| is the number of arrangements where P and R are together. Fix Q's position. Then, P and R can be seated together in two ways: either on the left and right of Q, or on the other side? Wait, no. If Q is fixed, there are two seats adjacent to Q. Let me visualize the table: Q is fixed at position 1. The other seats are 2, 3, 4. Seat 2 is adjacent to Q (seat 1), seat 4 is adjacent to Q. Seat 3 is opposite Q. If P and R are to be seated together, they can be in seats 2 and 3, or seats 3 and 4, or seats 4 and 2? Wait, no. If P and R are to be adjacent, they need to be in seats that are next to each other. In a circular table with Q fixed at seat 1, the adjacent seats are 2 and 4. Seat 3 is opposite. So the possible adjacent pairs are (2,3), (3,4), (4,2). Wait, but seat 3 is not adjacent to seat 2 or 4? No, in a circular table of 4 seats, each seat is adjacent to two others. So seat 1 (Q) is adjacent to 2 and 4. Seat 2 is adjacent to 1 and 3. Seat 3 is adjacent to 2 and 4. Seat 4 is adjacent to 3 and 1. So the adjacent pairs are (1,2), (2,3), (3,4), (4,1). So if Q is fixed at seat 1, then the adjacent pairs involving Q are (1,2) and (1,4). The other adjacent pairs are (2,3) and (3,4). So if we want P and R to be adjacent, they can be in (2,3), (3,4), or (1,2) or (1,4). But since Q is fixed at seat 1, if P and R are in (1,2), then one of them is Q? No, Q is fixed at seat 1, so P and R cannot be in seat 1. Therefore, P and R can only be in (2,3) or (3,4). So two possible adjacent pairs. For each of these, P and R can be arranged in 2 ways. Then, the remaining seat (seat 4 or seat 2) is occupied by T. So total number of arrangements where P and R are together is 2 (adjacent pairs) * 2 (arrangements of P and R) * 1 (T in the remaining seat) = 4. But earlier I thought it was 2. So my previous calculation was wrong. Therefore, |A| = 4, not 2. Similarly for |B| and |C|. So |A| + |B| + |C| = 3 * 4 = 12. But the total number of circular permutations is 6, which is impossible. Wait, this suggests another error.

Wait, no. If Q is fixed, the total number of arrangements is 3! = 6. If |A| is the number of arrangements where P and R are together, which we calculated as 4, but that's more than the total number of arrangements. Clearly, something is wrong.

Wait, let me list all possible arrangements with Q fixed at seat 1. The other seats are 2, 3, 4. The people are P, R, T. The possible arrangements are:

1. P, R, T in seats 2, 3, 4: P in 2, R in 3, T in 4

2. P, R, T in seats 2, 3, 4: P in 2, T in 3, R in 4

3. P, R, T in seats 2, 3, 4: R in 2, P in 3, T in 4

4. P, R, T in seats 2, 3, 4: R in 2, T in 3, P in 4

5. P, R, T in seats 2, 3, 4: T in 2, P in 3, R in 4

6. P, R, T in seats 2, 3, 4: T in 2, R in 3, P in 4

Now, which of these have P and R adjacent? Let's check:

1. P in 2, R in 3: adjacent (seats 2 and 3 are adjacent). Yes.

2. P in 2, T in 3, R in 4: P and R are not adjacent (seat 2 and 4 are not adjacent). No.

3. R in 2, P in 3: adjacent. Yes.

4. R in 2, T in 3, P in 4: R and P are not adjacent. No.

5. T in 2, P in 3, R in 4: P and R are not adjacent. No.

6. T in 2, R in 3, P in 4: R and P are not adjacent. No.

So only arrangements 1 and 3 have P and R adjacent. So |A| = 2. Similarly, |B| (P and T adjacent) and |C| (R and T adjacent) would also be 2 each. So total |A| + |B| + |C| = 6. Now, |A ∩ B| is the number of arrangements where P and R are together AND P and T are together, which means all three are together. How many such arrangements are there? From the list above, arrangements 1 and 3 have P and R together. Let's check if they also have P and T together.

Arrangement 1: P in 2, R in 3, T in 4. P is adjacent to R (seats 2-3) and T is in seat 4, which is adjacent to seat 3 (R) and seat 1 (Q). So P is not adjacent to T. So no.

Arrangement 3: R in 2, P in 3, T in 4. P is adjacent to R (seats 2-3) and T is in seat 4, adjacent to P (seat 3-4). So P is adjacent to T. So this arrangement has P adjacent to both R and T. So only arrangement 3 has all three together. Similarly, if we check for other intersections, like |A ∩ C| (P and R together and R and T together), which would also require all three together. Let's see: in arrangement 3, R is adjacent to P and T, so it's included. Are there others? Let's check arrangement 6: T in 2, R in 3, P in 4. R is adjacent to T (seats 2-3) and P is in seat 4, adjacent to R (seat 3-4). So this arrangement also has R adjacent to both T and P. So this is another arrangement where all three are together. Wait, but in our list above, arrangement 6 is T in 2, R in 3, P in 4. So R is between T and P, so all three are adjacent. So this is another arrangement where all three are together. Similarly, arrangement 5: T in 2, P in 3, R in 4. P is adjacent to T (seats 2-3) and R is in seat 4, adjacent to P (seat 3-4). So this also has all three together. Wait, but earlier I thought only arrangement 3 had all three together. Let me check:

Arrangement 1: P, R, T in seats 2, 3, 4: P in 2, R in 3, T in 4. P is adjacent to R, R is adjacent to T, but P is not adjacent to T. So only two adjacent pairs.

Arrangement 2: P in 2, T in 3, R in 4: P and T are adjacent (seats 2-3), T and R are adjacent (seats 3-4), but P and R are not adjacent.

Arrangement 3: R in 2, P in 3, T in 4: R and P are adjacent, P and T are adjacent, R and T are not adjacent.

Arrangement 4: R in 2, T in 3, P in 4: R and T are adjacent, T and P are adjacent, R and P are not adjacent.

Arrangement 5: T in 2, P in 3, R in 4: T and P are adjacent, P and R are adjacent, T and R are not adjacent.

Arrangement 6: T in 2, R in 3, P in 4: T and R are adjacent, R and P are adjacent, T and P are not adjacent.

So actually, there are 3 arrangements where all three are together: arrangements 3, 5, 6? Wait, no. Wait, arrangement 3: R, P, T in seats 2, 3, 4. P is between R and T, so all three are adjacent. Similarly, arrangement 5: T, P, R in seats 2, 3, 4. P is between T and R, so all three are adjacent. Arrangement 6: T, R, P in seats 2, 3, 4. R is between T and P, so all three are adjacent. So there are 3 arrangements where all three are together. Therefore, |A ∩ B| = |A ∩ C| = |B ∩ C| = 3? Wait, no. Let me think.

|A ∩ B| is the number of arrangements where P and R are together AND P and T are together. From the list, only arrangement 3 has P and R together and P and T together. Similarly, |A ∩ C| is the number of arrangements where P and R are together AND R and T are together. Which would be arrangements 3 and 6? Wait, arrangement 3 has P and R together and R and T not together? No, in arrangement 3, R is in 2, P in 3, T in 4. R is adjacent to P (seats 2-3), and T is in 4, adjacent to P (seats 3-4). So R is not adjacent to T. So only P and R are together, and P and T are together. So arrangement 3 has both P and R together and P and T together. Similarly, arrangement 5: T in 2, P in 3, R in 4. P is adjacent to T and R, so this has P and T together and P and R together. So this is also in |A ∩ B|. Similarly, arrangement 6: T in 2, R in 3, P in 4. R is adjacent to T and P, so this is in |B ∩ C|. So |A ∩ B| includes arrangements 3 and 5? Wait, no. Let me clarify:

|A ∩ B| is arrangements where P and R are together AND P and T are together. So in arrangement 3: P and R are together, P and T are together. In arrangement 5: P and T are together, P and R are together. So both arrangements 3 and 5 are in |A ∩ B|. Similarly, arrangement 6 is in |B ∩ C| (R and T together, R and P together). So |A ∩ B| has 2 arrangements, |A ∩ C| has 2 arrangements, |B ∩ C| has 2 arrangements. And |A ∩ B ∩ C| is the number of arrangements where all three pairs are together, which is all three together. How many such arrangements are there? From the list, arrangements 3, 5, 6 have all three together? No, in arrangement 3, P is between R and T, so all three are together. In arrangement 5, P is between T and R, so all three are together. In arrangement 6, R is between T and P, so all three are together. So there are 3 arrangements where all three are together. Therefore, |A ∩ B ∩ C| = 3.

So now, applying inclusion-exclusion:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| = 2 + 2 + 2 - 2 - 2 - 2 + 3 = 6 - 6 + 3 = 3. But the total number of arrangements is 6, so the number of valid arrangements is 6 - 3 = 3, which is wrong because we know it should be 0. Wait, this is still inconsistent. What's the issue?

Wait, the valid arrangements are those where no two of P, R, T are adjacent. But in n = 4, it's impossible, so the number should be 0. But according to this, it's 3. Clearly, something is wrong. Let me check the calculations again.

|A| = 2, |B| = 2, |C| = 2. Total 6.

|A ∩ B| = 2 (arrangements 3 and 5), |A ∩ C| = 2 (arrangements 3 and 6?), |B ∩ C| = 2 (arrangements 6 and 5?).

Wait, let's define:

|A ∩ B|: arrangements where P and R are together AND P and T are together. These are arrangements where all three are together. From the list, arrangements 3 and 5 have P and R together and P and T together. Wait, arrangement 3: P is between R and T, so P is adjacent to both. Arrangement 5: P is between T and R, same thing. So |A ∩ B| = 2.

Similarly, |A ∩ C|: arrangements where P and R are together AND R and T are together. These are arrangements where all three are together. From the list, arrangement 3: R is between P and T, so yes. Arrangement 6: R is between T and P, so yes. So |A ∩ C| = 2.

Similarly, |B ∩ C|: arrangements where P and T are together AND R and T are together. These are arrangements where all three are together. From the list, arrangement 5: T is between P and R, and arrangement 6: T is between R and P. Wait, arrangement 5: T in 2, P in 3, R in 4. T is adjacent to P and R. Arrangement 6: T in 2, R in 3, P in 4. T is adjacent to R and P. So |B ∩ C| = 2.

Therefore, |A ∩ B| = |A ∩ C| = |B ∩ C| = 2.

And |A ∩ B ∩ C| is the number of arrangements where all three pairs are together, which is all three together. From the list, there are 3 such arrangements: 3, 5, 6. Wait, but earlier I thought |A ∩ B| = 2, but if |A ∩ B ∩ C| is 3, then how?

Wait, no. |A ∩ B ∩ C| is the set of arrangements that are in all three intersections, i.e., where P and R are together, P and T are together, and R and T are together. Which is the same as all three being together. How many such arrangements are there? In the list, arrangements 3, 5, 6 all have all three together. So |A ∩ B ∩ C| = 3.

Therefore, applying inclusion-exclusion:

|A ∪ B ∪ C| = 2 + 2 + 2 - 2 - 2 - 2 + 3 = 6 - 6 + 3 = 3. But the total number of arrangements is 6, so the number of valid arrangements is 6 - 3 = 3, which is wrong. But we know it should be 0. Where is the mistake?

Ah! The mistake is in the calculation of |A|, |B|, |C|. Earlier, I said |A| = 2, but actually, in the list, there are 2 arrangements where P and R are together, but there are also arrangements where P and R are not together. But the total number of arrangements is 6, and |A ∪ B ∪ C| = 3, which would mean that 3 arrangements have at least one pair together, and 3 have none. But we know that in n = 4, it's impossible to have none. So clearly, there is a mistake in the calculation of |A|, |B|, |C|.

Wait, let me re-examine |A|. |A| is the number of arrangements where P and R are together. From the list, arrangements 1 and 3 have P and R together. So |A| = 2. Similarly, |B| (P and T together) is arrangements 2, 5. Wait, arrangement 2: P in 2, T in 3, R in 4. P and T are adjacent. Arrangement 5: T in 2, P in 3, R in 4. P and T are adjacent. So |B| = 2. Similarly, |C| (R and T together) is arrangements 2, 4, 6? Wait, arrangement 2: R in 4, T in 3, so R and T are adjacent. Arrangement 4: R in 2, T in 3, so R and T are adjacent. Arrangement 6: R in 3, T in 2, so R and T are adjacent. So |C| = 3? Wait, no. Let me check:

Arrangement 1: P, R, T in seats 2, 3, 4: P in 2, R in 3, T in 4. R and T are adjacent (seats 3-4). So this is in |C|.

Arrangement 2: P in 2, T in 3, R in 4: R and T are adjacent (seats 3-4). So in |C|.

Arrangement 3: R in 2, P in 3, T in 4: R and T are not adjacent (seats 2 and 4 are not adjacent). So not in |C|.

Arrangement 4: R in 2, T in 3, P in 4: R and T are adjacent (seats 2-3). So in |C|.

Arrangement 5: T in 2, P in 3, R in 4: R and T are not adjacent (seats 2 and 4 are not adjacent). So not in |C|.

Arrangement 6: T in 2, R in 3, P in 4: R and T are adjacent (seats 2-3). So in |C|.

So |C| has arrangements 1, 2, 4, 6. Wait, that's 4 arrangements. But total number of arrangements is 6. This suggests that |C| = 4, which contradicts earlier calculations. This is getting too confusing. It seems that my initial approach to calculate |A|, |B|, |C| is error-prone.

Given the time I've spent and the confusion, perhaps I should trust the gap method result for the original problem, which gave 144, and see if that makes sense.

If the gap method gives 144, and the total number of circular permutations is 720, then the number of invalid arrangements is 720 - 144 = 576. Let me see if this aligns with inclusion-exclusion.

If I assume that the number of valid arrangements is 144, then the number of invalid arrangements is 576. Let me try to calculate |A ∪ B ∪ C| = 576. Then, using inclusion-exclusion:

|A ∪ B ∪ C| = 3*|A| - 3*|A ∩ B| + |A ∩ B ∩ C| = 576.

But I need to find |A|, |A ∩ B|, etc. Let me try to calculate |A| again for the original problem.

For the original problem, n = 7, and |A| is the number of arrangements where P and R are together. Using the gap method, if we treat P and R as a block, then we have 6 entities: the block and the other 5 people. Number of circular arrangements is (6 - 1)! = 5! = 120. Multiply by 2 (for P-R or R-P) gives 240. So |A| = 240. Similarly for |B| and |C|. So |A| + |B| + |C| = 720.

Now, |A ∩ B| is the number of arrangements where P and R are together AND P and T are together, which means all three are together. As before, treating them as a block, number of circular arrangements is (5 - 1)! = 4! = 24. Multiply by 3! = 6 for internal arrangements, giving 24 * 6 = 144. So |A ∩ B| = 144. Similarly for |A ∩ C| and |B ∩ C|. So |A ∩ B| + |A ∩ C| + |B ∩ C| = 3 * 144 = 432.

Now, |A ∩ B ∩ C| is the number of arrangements where all three are together, which is the same as |A ∩ B|, which is 144. Wait, no. |A ∩ B ∩ C| is the number of arrangements where all three pairs are together, which is the same as all three being together, which is 144. So applying inclusion-exclusion:

|A ∪ B ∪ C| = 720 - 432 + 144 = 720 - 432 = 288 + 144 = 432. Wait, no: the formula is |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C| = 720 - 432 + 144 = 432. Therefore, the number of valid arrangements is 720 - 432 = 288. But this contradicts the gap method result of 144. So which is correct?

Wait, let me think. If the gap method gives 144, and inclusion-exclusion gives 288, there is a conflict. Let me check with a smaller case where I can calculate manually.

Let me take n = 5, with 3 people P, R, T, and 2 others, say Q and S. Let's calculate the number of valid arrangements using both methods.

Gap method: arrange the other 2 people (Q and S) around the table. Number of circular arrangements: (2 - 1)! = 1. Number of gaps: 2. Need to place P, R, T into the gaps, no two in the same gap. But there are 2 gaps and 3 people, which is impossible. So number of valid arrangements is 0. But let's check with inclusion-exclusion.

Total number of circular permutations: (5 - 1)! = 24.

Number of invalid arrangements: |A ∪ B ∪ C|.

|A| = |B| = |C| = 2 * (5 - 2)! = 2 * 6 = 12. So 3 * 12 = 36.

|A ∩ B| = number of arrangements where all three are together: (5 - 3)! * 3! = 2! * 6 = 12. So 3 * 12 = 36.

|A ∩ B ∩ C| = 12.

So |A ∪ B ∪ C| = 36 - 36 + 12 = 12. Therefore, valid arrangements = 24 - 12 = 12. But according to the gap method, it should be 0. Contradiction again. So clearly, there is a mistake in the inclusion-exclusion approach.

This suggests that the inclusion-exclusion approach is not being applied correctly, likely due to incorrect calculation of |A|, |A ∩ B|, etc. Given the time I've spent and the confusion, I think I should trust the gap method, which gave 144 for the original problem, and see if that makes sense.

In the original problem, arranging 4 people around the table: (4 - 1)! = 6. Number of gaps: 4. Choose 3 gaps: C(4, 3) = 4. Arrange P, R, T: 3! = 6. Total: 6 * 4 * 6 = 144. This seems logical. Let me verify with n = 5, where the gap method would give: arrange 2 people, (2 - 1)! = 1. Gaps: 2. Need to place 3 people into 2 gaps: impossible. So 0, which matches the manual calculation. For n = 6, with 3 people to place, let's see: arrange 3 people around the table: (3 - 1)! = 2. Gaps: 3. Choose 3 gaps: C(3, 3) = 1. Arrange P, R, T: 3! = 6. Total: 2 * 1 * 6 = 12. Let's check with inclusion-exclusion.

Total circular permutations: (6 - 1)! = 120.

|A| = 2 * (6 - 2)! = 2 * 24 = 48. 3 * 48 = 144.

|A ∩ B| = (6 - 3)! * 3! = 6 * 6 = 36. 3 * 36 = 108.

|A ∩ B ∩ C| = 36.

So |A ∪ B ∪ C| = 144 - 108 + 36 = 72. Valid arrangements = 120 - 72 = 48. But according to the gap method, it's 12. Contradiction again. So clearly, the inclusion-exclusion approach is not being applied correctly.

Given this inconsistency, I think the gap method is more reliable here, especially since it works for the n = 4 and n = 5 cases. Therefore, I will go with the gap method result of 144 for the original problem.

But wait, let me think again. In the original problem, arranging 4 people around the table: (4 - 1)! = 6. Then, placing P, R, T in the gaps. But in circular arrangements, when you arrange the 4 people, the number of gaps is 4. Placing 3 people in 4 gaps, no two in the same gap: C(4, 3) * 3! = 4 * 6 = 24. Total arrangements: 6 * 24 = 144. This seems correct.

Alternatively, let me think of it as linear arrangements. If it were a linear arrangement, the number would be: arrange the 4 people in 4! ways, then choose 3 gaps out of 5 (since in linear, there are n + 1 gaps), but since it's circular, it's different. But the gap method for circular is to arrange the non-restricted people, then place the restricted ones in the gaps. This is a standard method, and I think it is correct.

Therefore, I will conclude that the number of valid arrangements is 144. But let me check with another approach.

Another way: total number of circular permutations: 6! = 720.

Number of ways where no two of P, R, T are adjacent: ?

First, fix P's position to remove rotational symmetry. Now, we need to arrange the remaining 6 people, but with the condition that R and T are not adjacent to P or each other.

Wait, fixing P's position, the table has 6 seats left. We need to place R and T such that they are not adjacent to P or to each other.

Wait, this might be complex, but let's try.

Fix P at a position. Now, the seats adjacent to P are two seats. We need to ensure that R and T are not in those two seats, and also not adjacent to each other.

So, first, choose seats for R and T. There are 6 seats left, excluding P's seat. The two seats adjacent to P are forbidden for R and T. So, available seats: 6 - 2 = 4 seats. Let me visualize: P is fixed, seats are labeled 1 (P), 2, 3, 4, 5, 6. Seats 2 and 6 are adjacent to P. So available seats for R and T are 3, 4, 5. Wait, no: total seats are 7, with P at seat 1. Seats 2 and 7 are adjacent to P. Wait, if it's a circular table, seat 1 is adjacent to 2 and 7. So available seats for R and T are seats 3, 4, 5, 6. Wait, 7 - 1 (P) - 2 (adjacent to P) = 4 seats. So seats 3, 4, 5, 6.

Now, we need to place R and T in these 4 seats such that they are not adjacent to each other. How many ways?

First, choose 2 seats out of 4 for R and T, such that they are not adjacent. Let's find the number of ways to choose 2 non-adjacent seats in a line of 4 seats (since seats 3, 4, 5, 6 are in a line? No, in a circular table, but since we've fixed P's position and excluded adjacent seats, the remaining seats are in a line? Wait, no. If P is fixed at seat 1, seats 2 and 7 are adjacent to P. The remaining seats are 3, 4, 5, 6. These are in a line? No, in a circular table, seat 3 is adjacent to 4 and 2, seat 4 is adjacent to 3 and 5, seat 5 is adjacent to 4 and 6, seat 6 is adjacent to 5 and 7. But seat 7 is adjacent to P (seat 1) and seat 6. But we have excluded seat 2 and 7 for R and T. Wait, no, we have excluded seats 2 and 7 (adjacent to P) for R and T. So R and T can be placed in seats 3, 4, 5, 6. Now, in this set of seats, seat 3 is adjacent to 4, seat 4 to 5, seat 5 to 6, and seat 6 is adjacent to 5 and 7 (but 7 is excluded). So in the available seats 3, 4, 5, 6, the adjacency is between 3-4, 4-5, 5-6. So it's like a line of 4 seats: 3-4-5-6. So the number of ways to choose 2 non-adjacent seats in a line of 4 seats is C(4, 2) - 3 = 6 - 3 = 3. Wait, in a line of 4 seats, the number of adjacent pairs is 3 (3-4, 4-5, 5-6). Total pairs: 6. Non-adjacent pairs: 6 - 3 = 3. These are (3,5), (3,6), (4,6). So 3 ways.

For each such choice, we can arrange R and T in 2! ways. So total ways: 3 * 2 = 6.

Now, after placing R and T, the remaining 4 people (Q, S, T, U? Wait, original problem has 7 people: P, R, T, and 4 others. Wait, in the original problem, there are 7 people: P, R, T, and 4 others. So after fixing P, we have 6 seats left, and we placed R and T in 2 of them, leaving 4 seats for the other 4 people. Wait, no: total people are 7. Fix P, then we have 6 people left: R, T, and 4 others. We placed R and T in 2 seats, so the remaining 4 seats are for the other 4 people. The number of ways to arrange them is 4!.

So total number of arrangements: 3 * 2 * 4! = 6 * 24 = 144. Which matches the gap method result! So this confirms that the number of valid arrangements is 144.

Therefore, the answer is 144. But wait, the question says "Seating arrangements which are rotations of each other are treated as the same." So we have already accounted for that by fixing P's position, or by using the gap method which inherently accounts for circular permutations. Therefore, the final answer is 144.

But let me check with the inclusion-exclusion result. Earlier, using inclusion-exclusion, I got 288, but that must have been incorrect. Now, with this method, it's 144. Since this method is verified with a smaller case, I trust it.

Therefore, the number of ways is 144, so the answer is \boxed{144}.

**Final Answer**
\boxed{144}

To determine the number of ways 7 people can sit around a round table such that **Pierre (P), Rosa (R), and Thomas (T)** are **not adjacent to each other**, we use a **gap method** that accounts for the circular nature of the table and the restriction on adjacency.

---

### Step 1: Total Circular Permutations

In a circular arrangement of $ n $ people, the number of distinct seating arrangements is:

$$
(n - 1)! = 6! = 720
$$

This is because rotations of the same arrangement are considered identical.

---

### Step 2: Apply the Gap Method

We want to **avoid** any two of P, R, and T sitting next to each other. A standard technique for such problems is to:

1. **Fix the positions of the other people** (those not in the restricted group).
2. **Place the restricted people (P, R, T)** in the **gaps** between the others, ensuring that no two of them are adjacent.

#### Step 2.1: Arrange the Other 4 People

We first arrange the 4 people who are **not** P, R, or T. Since the table is circular, the number of distinct arrangements is:

$$
(4 - 1)! = 3! = 6
$$

These 4 people create **4 gaps** between them (since it's a circle).

#### Step 2.2: Choose Gaps for P, R, T

We need to place P, R, and T into **3 of these 4 gaps**, ensuring that **no two of them are in the same gap** (to prevent adjacency). The number of ways to choose 3 gaps from 4 is:

$$
\binom{4}{3} = 4
$$

#### Step 2.3: Arrange P, R, T in the Chosen Gaps

Once the gaps are chosen, we can arrange P, R, and T in those 3 gaps in:

$$
3! = 6 \text{ ways}
$$

---

### Step 3: Total Valid Arrangements

Now, multiply all the components:

- Ways to arrange the 4 non-restricted people: $ 3! = 6 $
- Ways to choose 3 gaps from 4: $ \binom{4}{3} = 4 $
- Ways to arrange P, R, T in the chosen gaps: $ 3! = 6 $

$$
\text{Total valid arrangements} = 6 \times 4 \times 6 = 144
$$

---

### Final Answer

$$
\boxed{144}
$$

Solution:
After Pierre sits, we can place Rosa either two seats from Pierre (that is, with one seat between them) or three seats from Pierre. We tackle these two cases separately:

Case 1: Rosa is two seats from Pierre. There are $2$ such seats. For either of these, there are then four empty seats in a row, and one empty seat between Rosa and Pierre. Thomas can sit in either of the middle two of the four empty seats in a row. So, there are $2\cdot 2 = 4$ ways to seat Rosa and Thomas in this case. There are then $4$ seats left, which the others can take in $4! = 24$ ways. So, there are $4\cdot 24 = 96$ seatings in this case.

Case 2: Rosa is three seats from Pierre (that is, there are $2$ seats between them). There are $2$ such seats. Thomas can't sit in either of the $2$ seats directly between them, but after Rosa sits, there are $3$ empty seats in a row still, and Thomas can only sit in the middle seat of these three. Once again, there are $4$ empty seats remaining, and the $4$ remaining people can sit in them in $4! = 24$ ways. So, we have $2\cdot 24 = 48$ seatings in this case.

Putting our two cases together gives a total of $96+48 = \boxed{144}$ seatings.