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 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.

So, there are 7 people total. If I first arrange the other 4 people (since P, R, T are the ones with restrictions), then I can place P, R, T in the gaps between them so that they aren't adjacent. But since it's a circular table, arranging the 4 people would have (4-1)! = 6 ways. Wait, but if I fix one person's position to account for circular rotations, then arranging the rest. Let me clarify.

In circular permutations, fixing one person's position is a common technique. Let me fix one of the non-P, R, T people, say, person A, to a seat. Then arrange the remaining 3 non-P, R, T people around the table. Wait, but there are 4 non-P, R, T people? Wait, total people are 7: P, R, T, and 4 others. So yes, 4 others. So if I fix one of them, say, person A, then the remaining 3 can be arranged in 3! ways. So total arrangements for the 4 non-P, R, T people is (4-1)! = 6? Wait, no. Wait, if we fix one person's position, then the rest can be arranged in (n-1)! So if we fix one person, then the number of ways to arrange the rest is (n-1)! So for 4 people, fixing one, the rest can be arranged in 3! = 6 ways. So that's correct.

Once these 4 people are seated, there are gaps between them where we can place P, R, T. Since it's a circular table, fixing one person's position, the number of gaps between the 4 people is 4. Wait, if there are 4 people seated around the table, there are 4 gaps between them. So we need to place 3 people (P, R, T) into these gaps, with no more than one person per gap, to ensure that none of them are adjacent. Wait, but if we place each of P, R, T into separate gaps, then they won't be adjacent. So the number of ways to choose the gaps is C(4,3) and then arrange P, R, T in those gaps. So C(4,3) is 4, and then 3! ways to arrange them. So total ways would be 3! * 4 * 3! ?

Wait, let me check again. First, arranging the 4 non-P, R, T people: fixing one person, so 3! ways. Then, the number of gaps is 4. Choosing 3 gaps out of 4: C(4,3) = 4. Then arranging P, R, T in those gaps: 3! ways. So total arrangements would be 3! * 4 * 3! = 6 * 4 * 6 = 144. But wait, the total number of circular permutations without restrictions is (7-1)! = 720. So 144 is less than that. But is this the correct approach?

Wait, but hold on. Let me think again. When we fix one person's position, we are effectively converting the circular table into a linear arrangement with that person fixed. So arranging the other 3 non-P, R, T people around the table would be 3! ways. Then, between these 4 people (including the fixed one), there are 4 gaps. So inserting P, R, T into different gaps. So yes, that seems right. So total number of arrangements would be 3! * C(4,3) * 3! = 6 * 4 * 6 = 144. But let me check if there's another way to compute this.

Alternatively, maybe I can use inclusion-exclusion. Let me try that approach as well to verify.

Total number of circular permutations: (7-1)! = 720.

Now, subtract the number of arrangements where at least two of P, R, T are sitting together. But inclusion-exclusion can get complicated here.

First, 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 the total.

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 have the same value.

So compute |A|: number of arrangements where P and R are sitting together.

In circular permutations, treating P and R as a single entity. But since it's a circular table, the number of ways to arrange n entities is (n-1)! So if we consider P and R as a single entity, then we have 6 entities: the PR block, T, and the other 4 people. Wait, total people are 7. If we combine P and R into one entity, then we have 6 entities. But since it's a circular table, the number of arrangements is (6-1)! = 5! = 120. However, within the PR block, P and R can be arranged in 2 ways. So total |A| = 2 * 5! = 240. Wait, but hold on, is that correct?

Wait, but in circular permutations, when we have a block, the formula is (number of entities - 1)! multiplied by the arrangements within the block. So yes, if we have 6 entities, it's (6-1)! * 2 = 120 * 2 = 240. But wait, but the total number of circular permutations is 720, so |A| is 240? But that seems high. Wait, but if we have 7 people, the number of arrangements where P and R are sitting together is 2 * 5! = 240? Let me check with another approach.

Alternatively, fix P's position to account for circular rotations. Then R can be seated either to the left or right of P, so 2 choices. Then the remaining 5 people can be arranged in 5! ways. Wait, but fixing P's position, the total number of arrangements is 6! = 720? Wait, no. Wait, if we fix P's position, then the rest can be arranged in 6! ways? Wait, no. Wait, in circular permutations, fixing one person's position, the rest can be arranged in (n-1)! So for 7 people, fixing P's position, the rest can be arranged in 6! ways. But if we fix P's position, then R can be seated in 6 positions, but adjacent to P are two seats. Wait, but if we fix P's position, then there are 6 seats left. The number of arrangements where R is adjacent to P is 2 (left or right) * 5! (arrangements of the rest). Wait, but that would be 2 * 5! = 240. But total number of arrangements with P fixed is 6! = 720. Wait, but 2 * 5! = 240, which is exactly 1/3 of 720. That seems plausible? Wait, but if you fix P, the probability that R is adjacent to P is 2/6 = 1/3, so yes, 1/3 of the total arrangements. So that seems correct. Therefore, |A| = 240. Similarly, |B| and |C| are also 240 each. So |A| + |B| + |C| = 3 * 240 = 720.

But then we have to subtract |A ∩ B|, |A ∩ C|, |B ∩ C|. Let's compute |A ∩ B|: arrangements where both P and R are together and P and T are together. So that would mean that P is sitting between R and T, so all three P, R, T are sitting together. Wait, but if both P and R are together and P and T are together, then R and T are also together? Because if P is between R and T, then R and T are adjacent? Wait, not necessarily. Wait, suppose we have P and R together, and P and T together. So P is adjacent to both R and T. So that would mean that R and T are on either side of P, so they are adjacent to each other? Wait, no. If P is between R and T, then R and T are sitting next to each other? Wait, if you have a circular table, and P is sitting between R and T, then R and T are adjacent? Wait, no. If you have R-P-T, then R and T are not adjacent. Wait, but in a circular table, if you have R-P-T, then R is adjacent to P and T is adjacent to P, but R and T are not adjacent. Wait, but if you have a block of R-P-T, then R and T are separated by P. Wait, but if you have P adjacent to both R and T, then R and T are not adjacent. Wait, but if you have P and R together, and P and T together, then R and T could be on either side of P, so they are not adjacent. However, if you have P and R together, and P and T together, then the three of them form a block of size 3? Wait, no. If P is adjacent to both R and T, then R and T are on either side of P, so they are not adjacent. So the three of them are in a line: R-P-T, but since it's a circular table, but they are not necessarily forming a block. Wait, but if you have P adjacent to both R and T, then R and T are not adjacent. However, if you consider the entire arrangement, then the three people P, R, T are all adjacent to P, but not necessarily to each other. Wait, but in that case, how do we count the arrangements where both P and R are together and P and T are together?

Wait, maybe it's equivalent to having all three P, R, T together? Wait, no. Because if P is adjacent to both R and T, but R and T are not adjacent. Wait, but in a circular table, if you have three people where each is adjacent to P, but not necessarily to each other. Wait, but in reality, if you have P adjacent to R and P adjacent to T, then R and T are separated by P. So they are not adjacent. So in that case, the three people are not all together. However, if you have P, R, T all sitting together, then they form a block of three, and in that case, each of P, R, T is adjacent to two others. So in that case, both P and R are adjacent, P and T are adjacent, and R and T are adjacent. So the intersection of A and B (where A is P and R together, B is P and T together) includes both cases where all three are together and cases where only P is adjacent to both R and T but R and T are not adjacent. Wait, but is that possible?

Wait, suppose we have a circular table. Let me imagine seating. If P is between R and T, then R and T are not adjacent. So that's one case. But if all three are sitting together, then R and T are adjacent. So the intersection of A and B includes both possibilities. Therefore, to compute |A ∩ B|, we need to consider both cases. Wait, but how?

Alternatively, maybe treating A ∩ B as the arrangements where P is adjacent to both R and T. So how many such arrangements are there?

In circular permutations, if we fix P's position, then R and T can be seated in two seats adjacent to P. Since there are two seats adjacent to P, and we need to place R and T in those seats. So if we fix P's position, then there are 2 seats next to P. The number of ways to arrange R and T in those seats is 2! = 2. Then the remaining 4 people can be arranged in 4! ways. So total arrangements would be 2 * 4! = 48. But since we fixed P's position, the total number of arrangements is 6! = 720. Wait, but if we don't fix positions, how does that translate?

Wait, maybe another approach. If we consider the number of arrangements where P is adjacent to both R and T. In circular permutations, the number of ways to arrange the people so that P is adjacent to both R and T. Let me think of P as fixed. Then R and T can be on either side of P. So there are 2! ways to arrange R and T next to P. Then the remaining 4 people can be arranged in 4! ways. So total arrangements with P fixed is 2! * 4! = 48. Since fixing P's position accounts for circular rotations, the total number of such arrangements is 48. But if we don't fix positions, then the total number would be 48 * 7 / 7? Wait, no. Wait, fixing P's position is a way to account for circular permutations. So if we fix P's position, then the total number of arrangements where P is adjacent to both R and T is 2! * 4! = 48. Therefore, |A ∩ B| is 48? But earlier, when I thought of treating P, R, T as a block, that would be different.

Wait, but if we consider that when P is adjacent to both R and T, they can be arranged as R-P-T or T-P-R, but also, if they are in a block of three, then there are more possibilities. Wait, no. If we fix P's position, then R and T are on either side of P, so the block is R-P-T or T-P-R. But if they are in a block of three, then the block can be arranged in 3! ways, but since we fixed P's position, the block would be fixed? Wait, maybe not. Wait, if we don't fix positions, then the number of arrangements where P, R, T are all together is calculated as follows: treat them as a block, so we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5-1)! = 4! = 24. Then multiply by the number of ways to arrange P, R, T within the block: 3! = 6. So total is 24 * 6 = 144. But earlier, when we fixed P's position, we found 48 arrangements where P is adjacent to both R and T. But 48 is less than 144. So there's a discrepancy here. Which one is correct?

Wait, if we fix P's position, then the number of arrangements where P is adjacent to both R and T is 2 * 4! = 48. However, the number of arrangements where all three are together is calculated as follows: if we fix P's position, then the block can be placed in two ways: either the block is P-R-T or P-T-R, but also, the block could be placed such that the block is on the other side? Wait, no. If we fix P's position, then the block of three can be arranged in two ways: R-P-T or T-P-R, but also, the block could be placed such that the block is on the other side? Wait, no. If P is fixed, then the block can be placed in two different ways: R-P-T or T-P-R. However, if the block is placed such that the entire block is on one side, but since it's a circular table, fixing P's position, the rest of the table is linear. Wait, maybe not. Let me think again.

If we fix P's position, then the rest of the seats are in a line relative to P. So, if we want all three P, R, T to be together, then the block can be placed either to the left of P or to the right of P? Wait, no. Since it's a circular table, fixing P's position, the block can be arranged in two different ways: R-P-T or T-P-R, but also, the block could be placed such that the block is on the other side? Wait, no. If we fix P's position, then the block of three people must include P and two others. So if we fix P, then the block can be arranged as R-P-T or T-P-R, but also, the block could be placed such that the block is on the other side? Wait, no. If you fix P's position, then the block can be placed in two different ways: either R and T are on the left and right of P, or... Wait, no. If you fix P's position, then the block of three people must include P and two others. So the block can be arranged in two ways: R-P-T or T-P-R. However, if the block is placed such that the entire block is on one side, but since it's a circular table, once P is fixed, the rest of the seats are linear. Wait, maybe not. Let me think of the seats as positions around the table. Fix P at position 1. Then positions 2 and 7 are adjacent to P. If we want all three P, R, T to be together, then R and T must be in positions 2 and 7, but also, there's position 3, 4, 5, 6. Wait, no. If we want all three P, R, T to be together, then they need to occupy three consecutive seats. Since P is fixed at position 1, the block can be positions 1, 2, 3 or positions 1, 7, 6. So two possible blocks. Each block can have R and T arranged in 2 ways (R-P-T or T-P-R). Then the remaining 4 people can be arranged in 4! ways. So total arrangements would be 2 * 2 * 4! = 2 * 2 * 24 = 96. Wait, but earlier when I fixed P's position and considered only P adjacent to both R and T, I had 2 * 4! = 48. So there's a difference here. Which is correct?

Wait, if we fix P's position, and want all three P, R, T to be together, then there are two possible blocks (positions 1-2-3 or 1-7-6). Each block has P in the middle or at the end? Wait, no. If the block is positions 1, 2, 3, then P is at position 1, and the block is P, R, T. But if the block is positions 1, 7, 6, then P is at position 1, and the block is P, T, R or P, R, T? Wait, if the block is positions 1, 7, 6, then P is at position 1, and the other two positions are 7 and 6. So the block could be P, T, R or P, R, T. So for each of the two blocks (positions 1-2-3 and 1-7-6), there are 2 arrangements of R and T. So total of 2 blocks * 2 arrangements = 4. Then the remaining 4 people can be arranged in 4! ways. So total arrangements would be 4 * 4! = 4 * 24 = 96. Therefore, the number of arrangements where all three are together is 96. However, earlier, when I considered arrangements where P is adjacent to both R and T, I had 2 * 4! = 48. So that suggests that there are two different cases: when all three are together (which includes P adjacent to both R and T) and when P is adjacent to both R and T but R and T are not adjacent. Wait, but if P is adjacent to both R and T, then R and T are on either side of P, so they are not adjacent. Wait, but in the case where all three are together, R and T are adjacent to each other. So the intersection of A and B (arrangements where P and R are together and P and T are together) includes both cases where R and T are adjacent (all three together) and where R and T are not adjacent (P between R and T). Therefore, the total number of arrangements where P is adjacent to both R and T is 96 (all three together) plus the number of arrangements where P is between R and T but R and T are not adjacent. Wait, but how?

Wait, if we fix P's position, then the number of arrangements where P is adjacent to both R and T is 2 * 4! = 48. But according to the previous calculation, the number of arrangements where all three are together is 96. So there's a contradiction here. Which is correct?

Wait, maybe my first approach was wrong. Let me think again. If we fix P's position, then the number of ways that R and T are adjacent to P is 2 (positions left and right) * 2! (arrangements of R and T) * 4! (arrangements of the rest). Wait, no. If we fix P's position, then there are two seats adjacent to P. We need to place R and T in those two seats. The number of ways is 2! (arrangements of R and T) * 4! (arrangements of the other people). So that's 2 * 4! = 48. However, if we consider the case where all three are together, then we have more arrangements. Wait, but if all three are together, then they occupy three consecutive seats. If P is fixed, then the block can be in two positions: positions 1-2-3 or 1-7-6. In each of these blocks, P is at position 1, and the other two seats are occupied by R and T. However, in this case, the number of arrangements is 2 (positions for the block) * 2! (arrangements of R and T) * 4! (arrangements of the rest). Wait, that would be 2 * 2! * 4! = 2 * 2 * 24 = 96. But this seems to be double the previous number. Why?

Because when we fix P's position and consider the block of three, we have two different blocks (positions 1-2-3 and 1-7-6), each allowing R and T to be arranged in 2 ways. However, when we consider just P adjacent to both R and T, we are only considering the cases where R and T are on either side of P, which is only one block (positions 2 and 7). Wait, no. If we fix P at position 1, then the seats adjacent to P are 2 and 7. If we want P adjacent to both R and T, then R and T must be in seats 2 and 7. However, if we have a block of three people including P, R, T, then the block could be positions 1-2-3 or 1-7-6. In the first case, seat 2 is adjacent to P, and seat 3 is adjacent to seat 2. In the second case, seat 7 is adjacent to P, and seat 6 is adjacent to seat 7. So in both cases, the block includes P and two others, but in the first case, the block extends to the right of P, and in the second case, to the left. However, if we only consider P adjacent to both R and T, then R and T are in seats 2 and 7, and the rest of the people are arranged. However, if we have a block of three, then seat 3 or seat 6 is also occupied by someone else. Wait, no. If the block is positions 1-2-3, then seat 1 is P, seat 2 is R, seat 3 is T. Then seat 3 is adjacent to seat 2 (R) and seat 4. Similarly, seat 7 is adjacent to seat 6 and seat 1 (P). So if the block is positions 1-2-3, then T is adjacent to P and seat 3. But in this case, T is adjacent to P, and R is adjacent to P. So this is included in the case where P is adjacent to both R and T. Similarly, if the block is positions 1-7-6, then seat 7 is T and seat 6 is R, so T is adjacent to P and seat 6, and R is adjacent to P and seat 5. So again, P is adjacent to both R and T. Therefore, the number of arrangements where all three are together is included in the count of arrangements where P is adjacent to both R and T. Therefore, the count of 48 (from fixing P's position) includes both cases where all three are together and where they are not. Wait, but how?

Wait, if we fix P's position, and count the number of arrangements where R and T are adjacent to P, that is, R and T are in seats 2 and 7. Then, the rest of the people (the other 4) can be arranged in 4! ways. However, in this case, the block of three (P, R, T) is not necessarily all together. Wait, no. If R and T are in seats 2 and 7, then they are not adjacent to each other. So the three people P, R, T are not all together. However, if we have a block of three, then they are all together. Therefore, the count of 48 (arrangements where P is adjacent to both R and T) does not include the cases where all three are together. Wait, but how?

Wait, if we fix P's position, and have R and T in seats 2 and 7, then they are not adjacent. However, if we have a block of three, then R and T are adjacent to each other. Therefore, the number of arrangements where all three are together is different from the number of arrangements where P is adjacent to both R and T. Therefore, there are two different cases. Therefore, the intersection of A and B (arrangements where P and R are together and P and T are together) includes both cases where all three are together and where only P is adjacent to both R and T. Therefore, how do we calculate |A ∩ B|?

Wait, maybe I need to think of it as follows: the number of arrangements where P is adjacent to both R and T is equal to the number of arrangements where P is between R and T (so R and T are not adjacent) plus the number of arrangements where all three are together. Wait, but how do we calculate that?

Alternatively, maybe using the formula for the number of ways where two specific people are adjacent and another specific person is adjacent to the first. Wait, this is getting too convoluted. Maybe I need to use the inclusion-exclusion formula properly.

Alternatively, maybe I should use the formula for |A ∩ B|. Let me think of A as P and R together, B as P and T together. So |A ∩ B| is the number of arrangements where both P and R are together and P and T are together. Which, as discussed, can be calculated by considering that P is adjacent to both R and T. So in circular permutations, how many ways are there for P to be adjacent to both R and T?

If we consider P, R, T as a single entity where P is between R and T. Wait, but since it's a circular table, maybe we can think of it as arranging the block R-P-T or T-P-R. But since it's a circular table, fixing P's position might help. Wait, if we fix P's position, then R and T can be on either side of P. So there are 2! ways to arrange R and T next to P. Then the remaining 4 people can be arranged in 4! ways. So total arrangements with P fixed is 2! * 4! = 48. Since fixing P's position accounts for circular rotations, the total number of such arrangements is 48. Therefore, |A ∩ B| = 48. Similarly, |A ∩ C| and |B ∩ C| are also 48 each. Therefore, the total for |A ∩ B| + |A ∩ C| + |B ∩ C| is 3 * 48 = 144.

Now, moving on to |A ∩ B ∩ C|, which is the number of arrangements where all three pairs are together, i.e., all three P, R, T are together. How do we calculate that?

As before, treating P, R, T as a single block. Then, we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5-1)! = 4! = 24. Then, within the block, the number of arrangements is 3! = 6. So total is 24 * 6 = 144. However, if we fix P's position, how does that compare? If we fix P's position, then the block can be arranged in 2 ways (R-P-T or T-P-R) and the rest can be arranged in 4! ways. Wait, but earlier we had 2 * 4! = 48. But according to this, it's 144. There's a discrepancy here. Which is correct?

Wait, if we don't fix positions, the total number of circular arrangements where all three are together is 144. However, if we fix P's position, then the number of such arrangements is 2 * 4! = 48. Since fixing P's position reduces the total number of arrangements by a factor of 7 (since there are 7 positions). Wait, no. Wait, fixing P's position is a way to account for circular rotations. So if the total number without fixing is 144, then fixing P's position would give 144 / 7? No, that doesn't make sense. Wait, actually, when we fix P's position, we are effectively dividing by 7 (the number of rotations). So if the total number of arrangements where all three are together is 144, then fixing P's position would give 144 / 7? But that's not an integer. Wait, something is wrong here.

Wait, let me recast. The total number of circular arrangements where all three are together is calculated as follows: treat the block as a single entity, so we have 5 entities. The number of circular arrangements is (5-1)! = 24. Multiply by the number of ways to arrange the block: 3! = 6. So total is 24 * 6 = 144. However, if we fix P's position, then how many such arrangements are there? If we fix P's position, then the block can be arranged in two ways: R-P-T or T-P-R, and the rest of the people can be arranged in 4! ways. So that's 2 * 4! = 48. But since fixing P's position accounts for circular rotations, the total number of arrangements without fixing is 48 * 7? No, that would be overcounting. Wait, no. If we fix P's position, we are considering only arrangements where P is in a specific seat, so the total number of arrangements where all three are together is 48. But according to the previous calculation, it's 144. Therefore, there's a contradiction. Which is correct?

Wait, let's think of a smaller case. Suppose we have 3 people: P, R, T. How many circular arrangements are there where all three are together? Well, since it's a circular table, there's only 1 arrangement (since rotations are considered the same). But according to the formula: treat them as a block, so 1 entity, circular arrangements: (1-1)! = 1, multiplied by 3! = 6. Wait, that gives 6, which is wrong. Wait, clearly, the formula is wrong here. Wait, no. Wait, if we have 3 people, the number of circular arrangements is (3-1)! = 2. But if we treat them as a block, we have 1 entity, so (1-1)! = 1, multiplied by 3! = 6. Which is wrong. Therefore, my previous approach is flawed.

Ah! I see the mistake. When we treat the block as a single entity, the formula for circular permutations is (number of entities - 1)! So if we have a block of 3 people and 4 other people, total entities is 5, so (5-1)! = 24. But if we have only the block, then it's (1-1)! = 1. However, in reality, when we have a block of 3 people, the number of circular arrangements is (number of entities - 1)! multiplied by the arrangements within the block. Wait, but in the case of only the block, it's (1-1)! * 3! = 1 * 6 = 6, but the actual number of circular arrangements for 3 people is 2. Therefore, there's a problem here. So my previous approach is wrong.

Wait, so what's the correct way to calculate the number of circular arrangements where all three are together? Let me think again. If 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: treat the k people as a single block, so we have (n - k + 1) entities. The number of circular arrangements is ( (n - k + 1) - 1 )! = (n - k)! multiplied by the number of ways to arrange the k people within the block, which is k!. Therefore, total is (n - k)! * k!.

Wait, let's test this with n=3, k=3. Then it would be (3 - 3)! * 3! = 0! * 6 = 1 * 6 = 6. But the actual number of circular arrangements for 3 people is 2. So this formula is also wrong. Therefore, my approach is incorrect.

Wait, maybe the formula is different. Let me recall. For linear arrangements, the number of ways to have a block of k people is (n - k + 1)! * k!. For circular arrangements, it's different. Let me think. When arranging in a circle, fixing one person's position to remove rotational symmetry. Suppose we fix one person in the block, say P. Then the rest of the block can be arranged in (k - 1)! ways, and the remaining (n - k) people can be arranged in (n - k)! ways. Therefore, total number of circular arrangements is (k - 1)! * (n - k)!.

Wait, let's test this with n=3, k=3. Then it would be (3 - 1)! * (3 - 3)! = 2! * 0! = 2 * 1 = 2, which is correct. For n=4, k=2: number of circular arrangements where two specific people are together. Fix one person, say P, then the other person can be in 1 seat adjacent to P (since it's a circular table with 4 seats). Wait, no. If we fix P's position, the other person can be in 2 seats adjacent to P. Wait, but if we want them to be together, then the other person must be in one of the two seats adjacent to P. Then the remaining 2 people can be arranged in 2! ways. So total arrangements: 2 * 2! = 4. According to the formula: (k - 1)! * (n - k)! = (2 - 1)! * (4 - 2)! = 1! * 2! = 1 * 2 = 2. Which is different. So this formula is also wrong.

Hmm, this is getting too confusing. Maybe I need to find another way. Let me go back to the original problem.

Earlier, I had two different approaches: one where I calculated the valid arrangements by arranging the other people first and placing P, R, T in the gaps, getting 144, and another approach using inclusion-exclusion which is getting tangled up. Maybe I should trust the first method?

Let me think again. If I fix one of the non-P, R, T people, say person A, then arrange the remaining 3 non-P, R, T people around the table. That would be 3! ways. Then, there are 4 gaps between the 4 people (including person A). We need to place P, R, T into these gaps, one per gap, so that they are not adjacent. The number of ways to choose the gaps is C(4, 3) = 4, and then arrange P, R, T in those gaps: 3! ways. So total arrangements: 3! * 4 * 3! = 6 * 4 * 6 = 144.

But let me check with a smaller case. Suppose there are 4 people: P, R, T, and one other person, say A. How many ways can they sit around a table so that P, R, T are not all adjacent? Wait, but with 4 people, the condition would be that no two of P, R, T are adjacent. Wait, but with 4 people, if we have 3 people who can't be adjacent, it's impossible because there are only 4 seats. Wait, maybe not. Let me think. If we have 4 seats, and 3 people P, R, T who can't be adjacent. Wait, but in a circular table with 4 seats, each person has two neighbors. If we have 3 people, they must occupy 3 seats, so at least two of them will be adjacent. Therefore, the number of valid arrangements is zero. Let me check with the first method. Fix person A. Then arrange the remaining 0 non-P, R, T people (since total non-P, R, T is 1, which is A). So arranging 0 people: 1 way. Then, number of gaps is 1 (since there's only person A). Wait, no. If we fix person A, then there are 3 seats left. Wait, no. Wait, total people are 4: A, P, R, T. If we fix A's position, then the remaining 3 people (P, R, T) need to be arranged around the table. But the condition is that no two of P, R, T are adjacent. However, with 3 people in 3 seats around the table, they will all be adjacent to each other. Therefore, there are zero valid arrangements. According to the first method: arranging the non-P, R, T people (which is just A) with fixing A's position, so 0! = 1 way. Then, number of gaps is 1 (since there's only A). We need to place 3 people into 1 gap, which is impossible. Therefore, the number of ways is zero, which matches. So that seems correct.

Another test case: 5 people, with P, R, T needing to not sit next to each other. Let's see. Total people: 5. Non-P, R, T: 2 people, say A and B. Using the first method: fix A's position. Arrange the remaining 1 non-P, R, T person (B) in 1! = 1 way. Then, number of gaps is 2 (between A and B, and between B and A). Wait, no. If we have 2 people (A and B) fixed around the table, there are 2 gaps. We need to place P, R, T into these gaps, one per gap. But there are 2 gaps and 3 people, which is impossible. Therefore, the number of valid arrangements is zero. But is that true? Let's see. In a circular table with 5 seats, can we seat P, R, T such that none are adjacent? Let's try. If we have A and B seated, then the seats are A, ?, ?, ?, B. Wait, no. If we fix A's position, then B can be seated in one of the remaining seats. Wait, with 5 seats, fixing A's position, B can be in seat 2, 3, 4, or 5. Wait, but if we have two people (A and B), then there are 5 - 2 = 3 seats left for P, R, T. Wait, no. Wait, total people are 5: A, B, P, R, T. If we fix A's position, then B can be seated in any of the remaining 4 seats. Wait, but in the first method, I considered arranging the non-P, R, T people (A and B). So if we fix A's position, then B can be arranged in 1! = 1 way? Wait, no. Wait, if there are 2 non-P, R, T people (A and B), and we fix A's position, then B can be arranged in (2 - 1)! = 1! = 1 way? Wait, no. Wait, the formula for arranging n people around a table is (n-1)! So if we have 2 non-P, R, T people, fixing A's position, then B can be arranged in 1! = 1 way. So the two non-P, R, T people are A and B, fixed with A in one seat and B in another. Then, the number of gaps between them is 2. So we need to place P, R, T into these gaps. But there are 2 gaps and 3 people, which is impossible. Therefore, the number of valid arrangements is zero. But is that true? Let's see. If we have 5 seats, and we need to seat P, R, T such that none are adjacent. Let's try to place them. Suppose we have A and B seated with one seat between them. Then, the seats are A, X, B, Y, Z. Wait, but with 5 seats, if A and B are seated with one seat between them, then there are two seats between A and B. Wait, this is getting too confusing. Maybe it's better to accept that the first method works for the test case where it's impossible, giving zero, which is correct. So maybe the first method is correct.

Therefore, returning to the original problem, with 7 people, the first method gives 144. However, let me check with inclusion-exclusion. Earlier, I had:

Total arrangements: 720.

|A ∪ B ∪ C| = 3*240 - 3*48 + 144 = ?

Wait, earlier I had |A| = 240, |B| = 240, |C| = 240. Then |A ∩ B| = |A ∩ C| = |B ∩ C| = 48 each. And |A ∩ B ∩ C| = 144. Wait, but earlier I was confused about |A ∩ B ∩ C|. If I use the formula from the first method, where the number of arrangements where all three are together is 144, then:

|A ∪ B ∪ C| = 3*240 - 3*48 + 144 = 720 - 144 + 144 = 720? That can't be. Wait, no:

Wait, inclusion-exclusion formula is |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.

So plugging in:

= 3*240 - 3*48 + 144 = 720 - 144 + 144 = 720. Which would mean that |A ∪ B ∪ C| = 720, so the number of arrangements with no two of P, R, T together is total - |A ∪ B ∪ C| = 720 - 720 = 0, which is impossible. Clearly, something is wrong here.

This suggests that my calculation of |A ∩ B ∩ C| is wrong. If |A ∪ B ∪ C| = 720, then it would mean that all arrangements have at least two of P, R, T together, which is not true. For example, the first method suggests there are 144 valid arrangements. Therefore, my calculation of |A ∩ B ∩ C| is incorrect.

Let me try again. Let's recast the inclusion-exclusion.

First, |A|, |B|, |C|: each is the number of arrangements where a specific pair is together. Earlier, I thought |A| = 240. Let me verify this again.

Total number of circular arrangements: 720.

Number of arrangements where P and R are together: ?

As before, fixing P's position, R can be in 2 seats adjacent to P, and the rest can be arranged in 5! ways. Wait, no. If we fix P's position, then there are 6 seats left. R can be in 2 seats adjacent to P. Then the remaining 5 people can be arranged in 5! ways. So total arrangements with P and R together is 2 * 5! = 240. Since fixing P's position accounts for circular rotations, the total number is 240. Therefore, |A| = 240. Similarly for |B| and |C|.

Now, |A ∩ B|: arrangements where P and R are together AND P and T are together. As discussed earlier, this is equivalent to arrangements where P is adjacent to both R and T. How many are there?

If we fix P's position, then R and T must be in the two seats adjacent to P. The number of ways is 2! (arrangements of R and T) * 5! (arrangements of the rest)? Wait, no. If we fix P's position, then R and T are in the two seats adjacent to P, and the remaining 4 people can be arranged in 4! ways. Wait, total seats are 7. Fix P's position, then there are 6 seats left. If R and T are in two of those seats (adjacent to P), then the remaining 4 seats are for the other 4 people. So the number of arrangements is 2! * 4! = 2 * 24 = 48. Therefore, |A ∩ B| = 48. Similarly for |A ∩ C| and |B ∩ C|.

Now, |A ∩ B ∩ C|: arrangements where P and R are together, P and T are together, and R and T are together. This is the case where all three are together. How many such arrangements are there?

If we fix P's position, then R and T must be adjacent to P and also adjacent to each other. So the three of them form a block. As discussed earlier, there are two possible blocks: R-P-T or T-P-R. Then, the remaining 4 people can be arranged in 4! ways. So total arrangements with P fixed is 2 * 4! = 48. Therefore, the total number of such arrangements without fixing is 48 * 7 / 7? Wait, no. Wait, fixing P's position accounts for circular rotations, so the total number of arrangements where all three are together is 48. But earlier, when I thought of treating them as a block, I had a different number. Let me check with another approach.

If we treat P, R, T as a single block, then we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5-1)! = 24. Multiply by the number of ways to arrange the block: 3! = 6. So total is 24 * 6 = 144. But according to fixing P's position, it's 48. Which is correct?

Wait, if we don't fix positions, then the total number of circular arrangements where all three are together is 144. But if we fix P's position, it's 48. Since fixing P's position reduces the total number by a factor of 7 (number of seats), then 144 / 7 is not an integer, which is impossible. Therefore, there must be an error in one of the calculations.

Wait, let's think of it this way: when we fix P's position, the number of arrangements where all three are together is equal to the number of ways to arrange R and T next to P and also adjacent to each other. If P is fixed, then R and T must be on either side of P, forming a block of R-P-T or T-P-R. Then, the rest of the people can be arranged in 4! ways. So that's 2 * 4! = 48. Therefore, the total number of such arrangements without fixing is 48 * 7 / 7? No, fixing P's position already accounts for all rotations. Therefore, the total number of arrangements where all three are together is 48. But according to the block method, it's 144. There's a contradiction here.

Wait, let's think of the total number of circular arrangements where all three are together. If we have 7 people, and we want to count the number of circular arrangements where P, R, T are all together.

First, consider the number of ways to arrange the block of P, R, T and the other 4 people. In circular arrangements, the number is (number of entities - 1)! * arrangements within the block. Number of entities is 5 (block + 4 people). So (5-1)! = 24. Multiply by 3! = 6 for the block arrangements. Total is 24 * 6 = 144.

But if we fix P's position, then the number of such arrangements is 2 * 4! = 48. Since fixing P's position accounts for all rotations, the total number should be 48. But according to the block method, it's 144. This suggests that the block method is wrong.

Wait, but if we fix P's position, then the number of arrangements where all three are together is 48. Therefore, the total number of such arrangements without fixing is 48 * 7 / 7 = 48? No, that doesn't make sense. Wait, fixing P's position is just one specific arrangement. The total number of circular arrangements is 720. If we fix P's position, we're considering 720 / 7 = 102.857... which is not an integer. Wait, no. Wait, fixing P's position is a way to count the number of arrangements where P is in a specific seat. The total number of circular arrangements is (7-1)! = 720. If we fix P's position, the number of arrangements is 6! = 720. Wait, no! Wait, fixing P's position in a circular table reduces the problem to arranging the remaining 6 people linearly, which is 6! = 720. Wait, but that can't be, because the total number of circular arrangements is 720. So fixing P's position doesn't change the count; it's just a way to count them.

Wait, I think I'm getting confused between fixing a position and counting. Let me clarify:

Total number of circular arrangements: (7-1)! = 720.

If we fix P's position, then the number of arrangements is 6! = 720. Wait, that can't be. Wait, no. If you fix one person's position in a circular table, the number of arrangements is (n-1)! So for n=7, fixing P's position, the number of arrangements is 6! = 720? No, that's not possible. Wait, no. Wait, fixing one person's position in a circular table, the number of arrangements is (n-1)! So for n=7, it's 6! = 720. But the total number of circular arrangements is also 6! = 720. So fixing P's position is just a way to count them.

Therefore, if we fix P's position, and calculate the number of arrangements where all three are together, it's 2 * 4! = 48. Therefore, the total number of such arrangements without fixing is 48. But according to the block method, it's 144. This suggests that the block method is wrong.

Wait, but if we fix P's position, and calculate the number of arrangements where all three are together as 48, then the total number of such arrangements is 48. But if we don't fix P's position, then it should be 48 * 7 / 7 = 48? No, that doesn't make sense. Wait, no. If we fix P's position, we are considering all arrangements where P is in a specific seat. The total number of arrangements where all three are together is equal to the number of such arrangements with P fixed multiplied by 7 (number of seats P can be in)? No, because fixing P's position already accounts for all rotations.

This is very confusing. Maybe I need to find another way. Let me think of the number of ways where all three are together as follows:

First, choose the position of the block. In a circular table, the number of ways to place a block of 3 people is 7 (since it can start at any of the 7 seats). However, since it's a circular table, each block is counted 3 times (once for each seat in the block). Wait, no. For a block of 3 people, there are 7 possible starting positions, but each block is counted 3 times (once for each person in the block). Therefore, the number of distinct blocks is 7 / 3, which is not an integer. This suggests that my approach is wrong.

Alternatively, the number of ways to arrange the block is 7 (positions for the block) * 3! (arrangements within the block) * 4! (arrangements of the rest). But since it's a circular table, we need to divide by 7 to account for rotations. Wait, this is getting too tangled.

Maybe I should accept that my inclusion-exclusion approach is leading to contradictions and trust the first method which gave 144. Let me check with another approach.

The first method: arranging the 4 non-P, R, T people around the table, then placing P, R, T in the gaps.

Number of ways to arrange the 4 people: (4-1)! = 6.

Number of gaps: 4.

Choose 3 gaps: C(4,3) = 4.

Arrange P, R, T in the gaps: 3! = 6.

Total: 6 * 4 * 6 = 144.

This seems logical. Let me think of it as arranging the 4 people first, which creates 4 gaps. Placing P, R, T in separate gaps ensures they are not adjacent.

Alternatively, if I use the inclusion-exclusion, and assume that |A ∩ B ∩ C| is 48, then:

|A ∪ B ∪ C| = 3*240 - 3*48 + 48 = 720 - 144 + 48 = 624.

Then, the number of valid arrangements would be 720 - 624 = 96.

But this contradicts the first method's result of 144.

Alternatively, if |A ∩ B ∩ C| is 144, then:

|A ∪ B ∪ C| = 720 - 144 + 144 = 720, which is impossible.

Alternatively, if |A ∩ B ∩ C| is 24, then:

|A ∪ B ∪ C| = 720 - 144 + 24 = 504, and valid arrangements would be 720 - 504 = 216.

But this is just guessing.

Alternatively, let's think of |A ∩ B ∩ C| as the number of arrangements where all three are together, which is 144. But if we use the first method, the valid arrangements are 144, which would mean that the number of invalid arrangements is 720 - 144 = 576. But according to inclusion-exclusion, if |A ∪ B ∪ C| = 576, then:

576 = 3*240 - 3*|A ∩ B| + |A ∩ B ∩ C|.

So 576 = 720 - 3*|A ∩ B| + |A ∩ B ∩ C|.

Rearranged: 3*|A ∩ B| - |A ∩ B ∩ C| = 720 - 576 = 144.

If we assume |A ∩ B| = 48, then 3*48 - |A ∩ B ∩ C| = 144 => 144 - |A ∩ B ∩ C| = 144 => |A ∩ B ∩ C| = 0, which is impossible.

If we assume |A ∩ B| = 48 and |A ∩ B ∩ C| = 144, then 3*48 - 144 = 144 - 144 = 0 ≠ 144.

This is not working.

Maybe my calculation of |A ∩ B| is wrong. Let me think again.

If |A ∩ B| is the number of arrangements where P is adjacent to both R and T, which includes both cases where all three are together and where P is between R and T but R and T are not adjacent.

If I calculate this as follows:

Number of ways where P is adjacent to both R and T = number of ways where P is between R and T (R-P-T or T-P-R) + number of ways where all three are together.

Wait, but if all three are together, then P is between R and T. So actually, the number of arrangements where P is adjacent to both R and T is equal to the number of arrangements where all three are together. Wait, no. If all three are together, then P is adjacent to both R and T, but also R and T are adjacent. If P is between R and T but they are not together, then P is adjacent to both R and T, but R and T are not adjacent. So the total number of arrangements where P is adjacent to both R and T is equal to the number of arrangements where all three are together plus the number of arrangements where P is between R and T but R and T are not adjacent.

But how do we calculate this?

If we fix P's position, then the number of ways where P is adjacent to both R and T is 2 * 4! = 48, as before. This includes both cases where R and T are adjacent (all three together) and where they are not.

But how many of these are all three together? If we fix P's position, the number of arrangements where all three are together is 2 * 4! = 48? No, earlier we thought it was 2 * 4! = 48, but that seems to include both cases. Wait, no. If we fix P's position, and want all three to be together, then the block can be in two positions (to the left or right of P), and within the block, R and T can be arranged in 2 ways. So total is 2 * 2 * 4! = 96? No, if we fix P's position, and want all three together, then the block can be in two positions (positions 1-2-3 or 1-7-6), and within each block, R and T can be arranged in 2 ways. So total is 2 * 2 * 4! = 96? But this would mean that the number of arrangements where all three are together is 96, and the number of arrangements where P is adjacent to both R and T but they are not together is 48 - 96? No, this is getting too tangled.

At this point, I think I need to accept that my inclusion-exclusion approach is error-prone due to the complexity of overlapping cases, and trust the first method which gave 144.

But to verify, let's think of another way. The total number of ways where no two of P, R, T are adjacent is equal to the total number of circular permutations minus the number of permutations where at least two are adjacent.

But if the first method gives 144, and the total is 720, then the number of invalid arrangements is 720 - 144 = 576.

Alternatively, let's calculate it using the first method's logic.

Number of ways to arrange the 4 non-P, R, T people: (4-1)! = 6.

Number of ways to place P, R, T in the gaps: C(4,3) * 3! = 4 * 6 = 24.

Total: 6 * 24 = 144.

This seems logical.

Alternatively, if we use the formula for non-adjacent arrangements in circular tables.

For linear arrangements, the number of ways to arrange n people with k people not adjacent is C(n - k + 1, k) * k! * (n - k)! but for circular tables, it's different.

However, I recall that for circular arrangements, the formula is similar to linear but adjusted for circularity.

In this case, we have 7 seats. We want to place 3 people (P, R, T) such that no two are adjacent, and the rest are 4 people.

The formula for the number of ways to choose seats for P, R, T is C(n - k, k) + C(n - k - 1, k - 1) for circular tables? No, I need to recall.

Actually, the number of ways to choose k non-consecutive seats around a circular table of n seats is C(n - k, k) + C(n - k - 1, k - 1). But I'm not sure.

Alternatively, for circular tables, the number of ways to choose k seats such that no two are adjacent is C(n - k, k) + C(n - k - 1, k - 1).

But let me check with n=7, k=3.

C(7 - 3, 3) + C(7 - 3 - 1, 3 - 1) = C(4,3) + C(3,2) = 4 + 3 = 7.

But the number of ways to choose 3 non-consecutive seats around a circular table of 7 seats is 7? Let me think.

In a circular table with 7 seats, how many ways to choose 3 seats with no two adjacent?

This is equivalent to placing 3 people and 4 non-people around the table, no two people adjacent.

This is similar to the problem of arranging objects with spacing.

For circular arrangements, the formula is C(n - k, k) + C(n - k - 1, k - 1).

But if n=7, k=3, it gives 4 + 3 = 7. Let me verify.

Imagine the 7 seats. To choose 3 seats with no two adjacent.

This is equivalent to placing 3 people and 4 non-people around the table, with at least one non-person between each pair of people.

In circular arrangements, this is equal to (n - k) choose k divided by something?

Actually, the formula for the number of ways to choose k non-consecutive seats around a circular table of n seats is (n / (n - k)) * C(n - k, k). Wait, no.

Alternatively, it's C(n - k, k) + C(n - k - 1, k - 1). For n=7, k=3, it's C(4,3) + C(3,2) = 4 + 3 = 7.

Let me count manually.

In a circular table with 7 seats, how many ways to choose 3 seats with no two adjacent?

Let's fix one seat as seat 1.

If seat 1 is chosen, then seats 2 and 7 cannot be chosen. We need to choose 2 more seats from seats 3, 4, 5, 6, with no two adjacent.

This is equivalent to choosing 2 non-consecutive seats from 4 seats in a line (since seat 1 is fixed, and seats 2 and 7 are excluded). The number of ways is C(4 - 2 + 1, 2) = C(3, 2) = 3.

If seat 1 is not chosen, then we need to choose 3 seats from seats 2-7, with no two adjacent. This is equivalent to choosing 3 non-consecutive seats in a line of 6 seats, which is C(6 - 3 + 1, 3) = C(4, 3) = 4.

Wait, but this is for linear arrangements. For circular, if we fix seat 1 as not chosen, it becomes a linear arrangement. So total number is 3 + 4 = 7, which matches the formula.

Therefore, the number of ways to choose the seats is 7.

Once the seats are chosen, the number of ways to arrange P, R, T is 3! = 6.

The number of ways to arrange the other 4 people is (4-1)! = 6 (since it's a circular table). Wait, no. Wait, if we have already fixed the seats for P, R, T, then the remaining 4 people can be arranged in 4! ways, but since the table is circular, we need to adjust for rotations. Wait, no. Wait, if we have already fixed the positions of P, R, T, then the remaining 4 people can be arranged in 4! ways, but since the table is circular, but we have already fixed the positions of P, R, T, which might have fixed the rotation.

Wait, this is getting complicated. Let me think.

If we first choose the seats for P, R, T, which are 3 specific seats with no two adjacent. Once these seats are chosen, the rest of the seats are for the other 4 people. Since the table is circular, but we have already fixed the positions of P, R, T, the remaining people can be arranged in 4! ways. However, since the table is circular, but we have already fixed the positions of P, R, T, which might have fixed the rotation.

Wait, no. If we choose the seats for P, R, T, then the arrangement is determined up to rotation. But since we have already chosen specific seats, the rotations are already accounted for.

Wait, this is confusing. Let me think of it as follows:

The total number of ways to arrange the 7 people with P, R, T not adjacent is equal to the number of ways to choose the seats for P, R, T (7 ways) multiplied by the number of ways to arrange P, R, T in those seats (3!) multiplied by the number of ways to arrange the other 4 people in the remaining seats (4!).

But since it's a circular table, we need to divide by the number of rotations? No, because we have already fixed the seats by choosing specific positions.

Wait, no. If we choose specific seats, then the number of arrangements is 7 (choices of seats) * 3! * 4! = 7 * 6 * 24 = 1008. But this is greater than the total number of circular arrangements (720), which is impossible.

Clearly, this approach is wrong.

The error is that choosing specific seats in a circular table already accounts for rotations. Wait, no. If we count the number of ways to choose seats as 7, and then multiply by arrangements, we are overcounting.

Actually, the correct way is:

The number of ways to arrange the 7 people with P, R, T not adjacent is equal to (number of ways to choose seats for P, R, T) * (number of ways to arrange P, R, T) * (number of ways to arrange the others).

But in circular arrangements, the number of ways to choose seats for P, R, T is 7 (as calculated earlier). However, this 7 already includes rotations. Wait, no. If we fix the table, the number of ways to choose 3 non-adjacent seats is 7, as calculated. Then, for each such choice, we can arrange P, R, T in 3! ways, and the others in 4! ways. But since the table is circular, we need to divide by 7 to account for rotations? No, because we have already fixed the seats.

Wait, this is very confusing. Let me think of it as linear arrangements.

If it were a linear table, the number of ways would be:

Number of ways to choose seats for P, R, T: C(n - k + 1, k) = C(7 - 3 + 1, 3) = C(5, 3) = 10.

Then, arrange P, R, T: 3! = 6.

Arrange the others: 4! = 24.

Total: 10 * 6 * 24 = 1440.

But for a circular table, it's different.

However, in our first method, we got 144, which is 1440 / 10 = 144. So perhaps for circular tables, we need to divide by n?

Wait, if for linear it's 1440, and for circular it's 1440 / 7 = 205.714... which is not an integer.

This suggests that this approach is also incorrect.

Given the time I've spent and the fact that the first method gave a logical result that passed the test case, I think I should go with the first method's answer of 144.

Therefore, the number of ways is 144.

But let me check with another approach.

The formula for the number of ways to arrange n people around a circular table with k people not adjacent is:

(n - k)! * C(n - k, k) + ...? No, I need to recall.

Alternatively, using the principle of inclusion-exclusion, but I need to resolve the earlier contradiction.

Let me try again with the inclusion-exclusion, assuming that |A ∩ B ∩ C| is 48.

So, |A ∪ B ∪ C| = 3*240 - 3*48 + 48 = 720 - 144 + 48 = 624.

Then, the number of valid arrangements is 720 - 624 = 96.

But this contradicts the first method's 144.

Alternatively, if |A ∩ B ∩ C| is 24, then:

|A ∪ B ∪ C| = 720 - 144 + 24 = 504, and valid arrangements = 720 - 504 = 216.

But how to determine |A ∩ B ∩ C|?

If we use the first method's result of 144 valid arrangements, then the invalid arrangements are 720 - 144 = 576.

So, 576 = |A ∪ B ∪ C| = 3*240 - 3*|A ∩ B| + |A ∩ B ∩ C|.

So, 576 = 720 - 3*|A ∩ B| + |A ∩ B ∩ C|.

Rearranged: 3*|A ∩ B| - |A ∩ B ∩ C| = 720 - 576 = 144.

If we assume that |A ∩ B| = 48, then 3*48 - |A ∩ B ∩ C| = 144 => 144 - |A ∩ B ∩ C| = 144 => |A ∩ B ∩ C| = 0, which is impossible.

If we assume |A ∩ B| = 56, then 3*56 - |A ∩ B ∩ C| = 144 => 168 - |A ∩ B ∩ C| = 144 => |A ∩ B ∩ C| = 24.

But how to find |A ∩ B|?

If |A ∩ B| is the number of arrangements where P is adjacent to both R and T, which we calculated as 48 when fixing P's position. If we don't fix P's position, it would be 48 * 7 / 7 = 48? No, this is not clear.

Alternatively, if |A ∩ B| is 48, and |A ∩ B ∩ C| is 24, then the equation holds.

But how to verify |A ∩ B ∩ C| = 24?

If we fix P's position, the number of arrangements where all three are together is 2 * 4! = 48. But if we don't fix P's position, it would be 48 * 7 / 7 = 48? No, this is not correct.

Alternatively, if the number of arrangements where all three are together is 144, then |A ∩ B ∩ C| = 144. Plugging into the equation: 3*|A ∩ B| - 144 = 144 => 3*|A ∩ B| = 288 => |A ∩ B| = 96.

But earlier, we calculated |A ∩ B| as 48.

This is very confusing.

Given the time I've spent and the fact that the first method gave a logical result, I think I should conclude that the correct answer is 144, and the inclusion-exclusion approach has errors in calculating the intersections.

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

**Final Answer**
\boxed{144}

To solve the problem of determining how many 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 proceed with a structured approach that avoids overcounting due to rotational symmetry and ensures the adjacency condition is respected.

---

### Step 1: Total Circular Permutations

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

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

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

---

### Step 2: Strategy to Avoid Adjacency

We want to **avoid** any arrangement where **P, R, and T are adjacent to each other**. A good strategy is to **first arrange the other 4 people** (let's call them A, B, C, D), and then **place P, R, T in the gaps** between them such that no two of them are adjacent.

This ensures that P, R, and T are **not adjacent** to each other.

---

### Step 3: Arrange the 4 Non-Adjacent People

We fix one of the 4 people (say, A) to eliminate rotational symmetry. The remaining 3 people (B, C, D) can be arranged in:

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

This gives us 6 distinct arrangements of the 4 people around the table.

---

### Step 4: Count the Gaps for P, R, T

When 4 people are seated around a circular table, they create **4 gaps** between them (one between each pair of adjacent people). These are the positions where we can place P, R, and T such that they are **not adjacent** to each other.

We need to choose **3 out of these 4 gaps** to place P, R, and T. The number of ways to choose 3 gaps from 4 is:

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

Then, we can arrange P, R, and T in these 3 chosen gaps in:

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

---

### Step 5: Total Valid Arrangements

Now, multiply the number of ways to arrange the 4 people, the number of ways to choose the gaps, and the number of ways to arrange P, R, and T in those gaps:

$$
\text{Total valid arrangements} = 3! \times \binom{4}{3} \times 3! = 6 \times 4 \times 6 = 144
$$

---

### 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 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.

So, there are 7 people total. If I first arrange the other 4 people (since P, R, T are the ones with restrictions), then I can place P, R, T in the gaps between them so that they aren't adjacent. But since it's a circular table, arranging the 4 people would have (4-1)! = 6 ways. Wait, but if I fix one person's position to account for circular rotations, then arranging the rest. Let me clarify.

In circular permutations, fixing one person's position is a common technique. Let me fix one of the non-P, R, T people, say, person A, to a seat. Then arrange the remaining 3 non-P, R, T people around the table. Wait, but there are 4 non-P, R, T people? Wait, total people are 7: P, R, T, and 4 others. So yes, 4 others. So if I fix one of them, say, person A, then the remaining 3 can be arranged in 3! ways. So total arrangements for the 4 non-P, R, T people is (4-1)! = 6? Wait, no. Wait, if we fix one person's position, then the rest can be arranged in (n-1)! So if we fix one person, then the number of ways to arrange the rest is (n-1)! So for 4 people, fixing one, the rest can be arranged in 3! = 6 ways. So that's correct.

Once these 4 people are seated, there are gaps between them where we can place P, R, T. Since it's a circular table, fixing one person's position, the number of gaps between the 4 people is 4. Wait, if there are 4 people seated around the table, there are 4 gaps between them. So we need to place 3 people (P, R, T) into these gaps, with no more than one person per gap, to ensure that none of them are adjacent. Wait, but if we place each of P, R, T into separate gaps, then they won't be adjacent. So the number of ways to choose the gaps is C(4,3) and then arrange P, R, T in those gaps. So C(4,3) is 4, and then 3! ways to arrange them. So total ways would be 3! * 4 * 3! ?

Wait, let me check again. First, arranging the 4 non-P, R, T people: fixing one person, so 3! ways. Then, the number of gaps is 4. Choosing 3 gaps out of 4: C(4,3) = 4. Then arranging P, R, T in those gaps: 3! ways. So total arrangements would be 3! * 4 * 3! = 6 * 4 * 6 = 144. But wait, the total number of circular permutations without restrictions is (7-1)! = 720. So 144 is less than that. But is this the correct approach?

Wait, but hold on. Let me think again. When we fix one person's position, we are effectively converting the circular table into a linear arrangement with that person fixed. So arranging the other 3 non-P, R, T people around the table would be 3! ways. Then, between these 4 people (including the fixed one), there are 4 gaps. So inserting P, R, T into different gaps. So yes, that seems right. So total number of arrangements would be 3! * C(4,3) * 3! = 6 * 4 * 6 = 144. But let me check if there's another way to compute this.

Alternatively, maybe I can use inclusion-exclusion. Let me try that approach as well to verify.

Total number of circular permutations: (7-1)! = 720.

Now, subtract the number of arrangements where at least two of P, R, T are sitting together. But inclusion-exclusion can get complicated here.

First, 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 the total.

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 have the same value.

So compute |A|: number of arrangements where P and R are sitting together.

In circular permutations, treating P and R as a single entity. But since it's a circular table, the number of ways to arrange n entities is (n-1)! So if we consider P and R as a single entity, then we have 6 entities: the PR block, T, and the other 4 people. Wait, total people are 7. If we combine P and R into one entity, then we have 6 entities. But since it's a circular table, the number of arrangements is (6-1)! = 5! = 120. However, within the PR block, P and R can be arranged in 2 ways. So total |A| = 2 * 5! = 240. Wait, but hold on, is that correct?

Wait, but in circular permutations, when we have a block, the formula is (number of entities - 1)! multiplied by the arrangements within the block. So yes, if we have 6 entities, it's (6-1)! * 2 = 120 * 2 = 240. But wait, but the total number of circular permutations is 720, so |A| is 240? But that seems high. Wait, but if we have 7 people, the number of arrangements where P and R are sitting together is 2 * 5! = 240? Let me check with another approach.

Alternatively, fix P's position to account for circular rotations. Then R can be seated either to the left or right of P, so 2 choices. Then the remaining 5 people can be arranged in 5! ways. Wait, but fixing P's position, the total number of arrangements is 6! = 720? Wait, no. Wait, if we fix P's position, then the rest can be arranged in 6! ways? Wait, no. Wait, in circular permutations, fixing one person's position, the rest can be arranged in (n-1)! So for 7 people, fixing P's position, the rest can be arranged in 6! ways. But if we fix P's position, then R can be seated in 6 positions, but adjacent to P are two seats. Wait, but if we fix P's position, then there are 6 seats left. The number of arrangements where R is adjacent to P is 2 (left or right) * 5! (arrangements of the rest). Wait, but that would be 2 * 5! = 240. But total number of arrangements with P fixed is 6! = 720. Wait, but 2 * 5! = 240, which is exactly 1/3 of 720. That seems plausible? Wait, but if you fix P, the probability that R is adjacent to P is 2/6 = 1/3, so yes, 1/3 of the total arrangements. So that seems correct. Therefore, |A| = 240. Similarly, |B| and |C| are also 240 each. So |A| + |B| + |C| = 3 * 240 = 720.

But then we have to subtract |A ∩ B|, |A ∩ C|, |B ∩ C|. Let's compute |A ∩ B|: arrangements where both P and R are together and P and T are together. So that would mean that P is sitting between R and T, so all three P, R, T are sitting together. Wait, but if both P and R are together and P and T are together, then R and T are also together? Because if P is between R and T, then R and T are adjacent? Wait, not necessarily. Wait, suppose we have P and R together, and P and T together. So P is adjacent to both R and T. So that would mean that R and T are on either side of P, so they are adjacent to each other? Wait, no. If P is between R and T, then R and T are sitting next to each other? Wait, if you have a circular table, and P is sitting between R and T, then R and T are adjacent? Wait, no. If you have R-P-T, then R and T are not adjacent. Wait, but in a circular table, if you have R-P-T, then R is adjacent to P and T is adjacent to P, but R and T are not adjacent. Wait, but if you have a block of R-P-T, then R and T are separated by P. Wait, but if you have P adjacent to both R and T, then R and T are not adjacent. Wait, but if you have P and R together, and P and T together, then R and T could be on either side of P, so they are not adjacent. However, if you have P and R together, and P and T together, then the three of them form a block of size 3? Wait, no. If P is adjacent to both R and T, then R and T are on either side of P, so they are not adjacent. So the three of them are in a line: R-P-T, but since it's a circular table, but they are not necessarily forming a block. Wait, but if you have P adjacent to both R and T, then R and T are not adjacent. However, if you consider the entire arrangement, then the three people P, R, T are all adjacent to P, but not necessarily to each other. Wait, but in that case, how do we count the arrangements where both P and R are together and P and T are together?

Wait, maybe it's equivalent to having all three P, R, T together? Wait, no. Because if P is adjacent to both R and T, but R and T are not adjacent. Wait, but in a circular table, if you have three people where each is adjacent to P, but not necessarily to each other. Wait, but in reality, if you have P adjacent to R and P adjacent to T, then R and T are separated by P. So they are not adjacent. So in that case, the three people are not all together. However, if you have P, R, T all sitting together, then they form a block of three, and in that case, each of P, R, T is adjacent to two others. So in that case, both P and R are adjacent, P and T are adjacent, and R and T are adjacent. So the intersection of A and B (where A is P and R together, B is P and T together) includes both cases where all three are together and cases where only P is adjacent to both R and T but R and T are not adjacent. Wait, but is that possible?

Wait, suppose we have a circular table. Let me imagine seating. If P is between R and T, then R and T are not adjacent. So that's one case. But if all three are sitting together, then R and T are adjacent. So the intersection of A and B includes both possibilities. Therefore, to compute |A ∩ B|, we need to consider both cases. Wait, but how?

Alternatively, maybe treating A ∩ B as the arrangements where P is adjacent to both R and T. So how many such arrangements are there?

In circular permutations, if we fix P's position, then R and T can be seated in two seats adjacent to P. Since there are two seats adjacent to P, and we need to place R and T in those seats. So if we fix P's position, then there are 2 seats next to P. The number of ways to arrange R and T in those seats is 2! = 2. Then the remaining 4 people can be arranged in 4! ways. So total arrangements would be 2 * 4! = 48. But since we fixed P's position, the total number of arrangements is 6! = 720. Wait, but if we don't fix positions, how does that translate?

Wait, maybe another approach. If we consider the number of arrangements where P is adjacent to both R and T. In circular permutations, the number of ways to arrange the people so that P is adjacent to both R and T. Let me think of P as fixed. Then R and T can be on either side of P. So there are 2! ways to arrange R and T next to P. Then the remaining 4 people can be arranged in 4! ways. So total arrangements with P fixed is 2! * 4! = 48. Since fixing P's position accounts for circular rotations, the total number of such arrangements is 48. But if we don't fix positions, then the total number would be 48 * 7 / 7? Wait, no. Wait, fixing P's position is a way to account for circular permutations. So if we fix P's position, then the total number of arrangements where P is adjacent to both R and T is 2! * 4! = 48. Therefore, |A ∩ B| is 48? But earlier, when I thought of treating P, R, T as a block, that would be different.

Wait, but if we consider that when P is adjacent to both R and T, they can be arranged as R-P-T or T-P-R, but also, if they are in a block of three, then there are more possibilities. Wait, no. If we fix P's position, then R and T are on either side of P, so the block is R-P-T or T-P-R. But if they are in a block of three, then the block can be arranged in 3! ways, but since we fixed P's position, the block would be fixed? Wait, maybe not. Wait, if we don't fix positions, then the number of arrangements where P, R, T are all together is calculated as follows: treat them as a block, so we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5-1)! = 4! = 24. Then multiply by the number of ways to arrange P, R, T within the block: 3! = 6. So total is 24 * 6 = 144. But earlier, when we fixed P's position, we found 48 arrangements where P is adjacent to both R and T. But 48 is less than 144. So there's a discrepancy here. Which one is correct?

Wait, if we fix P's position, then the number of arrangements where P is adjacent to both R and T is 2 * 4! = 48. However, the number of arrangements where all three are together is calculated as follows: if we fix P's position, then the block can be placed in two ways: either the block is P-R-T or P-T-R, but also, the block could be placed such that the block is on the other side? Wait, no. If we fix P's position, then the block of three can be arranged in two ways: R-P-T or T-P-R, but also, the block could be placed such that the block is on the other side? Wait, no. If P is fixed, then the block can be placed in two different ways: R-P-T or T-P-R. However, if the block is placed such that the entire block is on one side, but since it's a circular table, fixing P's position, the rest of the table is linear. Wait, maybe not. Let me think again.

If we fix P's position, then the rest of the seats are in a line relative to P. So, if we want all three P, R, T to be together, then the block can be placed either to the left of P or to the right of P? Wait, no. Since it's a circular table, fixing P's position, the block can be arranged in two different ways: R-P-T or T-P-R, but also, the block could be placed such that the block is on the other side? Wait, no. If we fix P's position, then the block of three people must include P and two others. So if we fix P, then the block can be arranged as R-P-T or T-P-R, but also, the block could be placed such that the block is on the other side? Wait, no. If you fix P's position, then the block can be placed in two different ways: either R and T are on the left and right of P, or... Wait, no. If you fix P's position, then the block of three people must include P and two others. So the block can be arranged in two ways: R-P-T or T-P-R. However, if the block is placed such that the entire block is on one side, but since it's a circular table, once P is fixed, the rest of the seats are linear. Wait, maybe not. Let me think of the seats as positions around the table. Fix P at position 1. Then positions 2 and 7 are adjacent to P. If we want all three P, R, T to be together, then R and T must be in positions 2 and 7, but also, there's position 3, 4, 5, 6. Wait, no. If we want all three P, R, T to be together, then they need to occupy three consecutive seats. Since P is fixed at position 1, the block can be positions 1, 2, 3 or positions 1, 7, 6. So two possible blocks. Each block can have R and T arranged in 2 ways (R-P-T or T-P-R). Then the remaining 4 people can be arranged in 4! ways. So total arrangements would be 2 * 2 * 4! = 2 * 2 * 24 = 96. Wait, but earlier when I fixed P's position and considered only P adjacent to both R and T, I had 2 * 4! = 48. So there's a difference here. Which is correct?

Wait, if we fix P's position, and want all three P, R, T to be together, then there are two possible blocks (positions 1-2-3 or 1-7-6). Each block has P in the middle or at the end? Wait, no. If the block is positions 1, 2, 3, then P is at position 1, and the block is P, R, T. But if the block is positions 1, 7, 6, then P is at position 1, and the block is P, T, R or P, R, T? Wait, if the block is positions 1, 7, 6, then P is at position 1, and the other two positions are 7 and 6. So the block could be P, T, R or P, R, T. So for each of the two blocks (positions 1-2-3 and 1-7-6), there are 2 arrangements of R and T. So total of 2 blocks * 2 arrangements = 4. Then the remaining 4 people can be arranged in 4! ways. So total arrangements would be 4 * 4! = 4 * 24 = 96. Therefore, the number of arrangements where all three are together is 96. However, earlier, when I considered arrangements where P is adjacent to both R and T, I had 2 * 4! = 48. So that suggests that there are two different cases: when all three are together (which includes P adjacent to both R and T) and when P is adjacent to both R and T but R and T are not adjacent. Wait, but if P is adjacent to both R and T, then R and T are on either side of P, so they are not adjacent. Wait, but in the case where all three are together, R and T are adjacent to each other. So the intersection of A and B (arrangements where P and R are together and P and T are together) includes both cases where R and T are adjacent (all three together) and where R and T are not adjacent (P between R and T). Therefore, the total number of arrangements where P is adjacent to both R and T is 96 (all three together) plus the number of arrangements where P is between R and T but R and T are not adjacent. Wait, but how?

Wait, if we fix P's position, then the number of arrangements where P is adjacent to both R and T is 2 * 4! = 48. But according to the previous calculation, the number of arrangements where all three are together is 96. So there's a contradiction here. Which is correct?

Wait, maybe my first approach was wrong. Let me think again. If we fix P's position, then the number of ways that R and T are adjacent to P is 2 (positions left and right) * 2! (arrangements of R and T) * 4! (arrangements of the rest). Wait, no. If we fix P's position, then there are two seats adjacent to P. We need to place R and T in those two seats. The number of ways is 2! (arrangements of R and T) * 4! (arrangements of the other people). So that's 2 * 4! = 48. However, if we consider the case where all three are together, then we have more arrangements. Wait, but if all three are together, then they occupy three consecutive seats. If P is fixed, then the block can be in two positions: positions 1-2-3 or 1-7-6. In each of these blocks, P is at position 1, and the other two seats are occupied by R and T. However, in this case, the number of arrangements is 2 (positions for the block) * 2! (arrangements of R and T) * 4! (arrangements of the rest). Wait, that would be 2 * 2! * 4! = 2 * 2 * 24 = 96. But this seems to be double the previous number. Why?

Because when we fix P's position and consider the block of three, we have two different blocks (positions 1-2-3 and 1-7-6), each allowing R and T to be arranged in 2 ways. However, when we consider just P adjacent to both R and T, we are only considering the cases where R and T are on either side of P, which is only one block (positions 2 and 7). Wait, no. If we fix P at position 1, then the seats adjacent to P are 2 and 7. If we want P adjacent to both R and T, then R and T must be in seats 2 and 7. However, if we have a block of three people including P, R, T, then the block could be positions 1-2-3 or 1-7-6. In the first case, seat 2 is adjacent to P, and seat 3 is adjacent to seat 2. In the second case, seat 7 is adjacent to P, and seat 6 is adjacent to seat 7. So in both cases, the block includes P and two others, but in the first case, the block extends to the right of P, and in the second case, to the left. However, if we only consider P adjacent to both R and T, then R and T are in seats 2 and 7, and the rest of the people are arranged. However, if we have a block of three, then seat 3 or seat 6 is also occupied by someone else. Wait, no. If the block is positions 1-2-3, then seat 1 is P, seat 2 is R, seat 3 is T. Then seat 3 is adjacent to seat 2 (R) and seat 4. Similarly, seat 7 is adjacent to seat 6 and seat 1 (P). So if the block is positions 1-2-3, then T is adjacent to P and seat 3. But in this case, T is adjacent to P, and R is adjacent to P. So this is included in the case where P is adjacent to both R and T. Similarly, if the block is positions 1-7-6, then seat 7 is T and seat 6 is R, so T is adjacent to P and seat 6, and R is adjacent to P and seat 5. So again, P is adjacent to both R and T. Therefore, the number of arrangements where all three are together is included in the count of arrangements where P is adjacent to both R and T. Therefore, the count of 48 (from fixing P's position) includes both cases where all three are together and where they are not. Wait, but how?

Wait, if we fix P's position, and count the number of arrangements where R and T are adjacent to P, that is, R and T are in seats 2 and 7. Then, the rest of the people (the other 4) can be arranged in 4! ways. However, in this case, the block of three (P, R, T) is not necessarily all together. Wait, no. If R and T are in seats 2 and 7, then they are not adjacent to each other. So the three people P, R, T are not all together. However, if we have a block of three, then they are all together. Therefore, the count of 48 (arrangements where P is adjacent to both R and T) does not include the cases where all three are together. Wait, but how?

Wait, if we fix P's position, and have R and T in seats 2 and 7, then they are not adjacent. However, if we have a block of three, then R and T are adjacent to each other. Therefore, the number of arrangements where all three are together is different from the number of arrangements where P is adjacent to both R and T. Therefore, there are two different cases. Therefore, the intersection of A and B (arrangements where P and R are together and P and T are together) includes both cases where all three are together and where only P is adjacent to both R and T. Therefore, how do we calculate |A ∩ B|?

Wait, maybe I need to think of it as follows: the number of arrangements where P is adjacent to both R and T is equal to the number of arrangements where P is between R and T (so R and T are not adjacent) plus the number of arrangements where all three are together. Wait, but how do we calculate that?

Alternatively, maybe using the formula for the number of ways where two specific people are adjacent and another specific person is adjacent to the first. Wait, this is getting too convoluted. Maybe I need to use the inclusion-exclusion formula properly.

Alternatively, maybe I should use the formula for |A ∩ B|. Let me think of A as P and R together, B as P and T together. So |A ∩ B| is the number of arrangements where both P and R are together and P and T are together. Which, as discussed, can be calculated by considering that P is adjacent to both R and T. So in circular permutations, how many ways are there for P to be adjacent to both R and T?

If we consider P, R, T as a single entity where P is between R and T. Wait, but since it's a circular table, maybe we can think of it as arranging the block R-P-T or T-P-R. But since it's a circular table, fixing P's position might help. Wait, if we fix P's position, then R and T can be on either side of P. So there are 2! ways to arrange R and T next to P. Then the remaining 4 people can be arranged in 4! ways. So total arrangements with P fixed is 2! * 4! = 48. Since fixing P's position accounts for circular rotations, the total number of such arrangements is 48. Therefore, |A ∩ B| = 48. Similarly, |A ∩ C| and |B ∩ C| are also 48 each. Therefore, the total for |A ∩ B| + |A ∩ C| + |B ∩ C| is 3 * 48 = 144.

Now, moving on to |A ∩ B ∩ C|, which is the number of arrangements where all three pairs are together, i.e., all three P, R, T are together. How do we calculate that?

As before, treating P, R, T as a single block. Then, we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5-1)! = 4! = 24. Then, within the block, the number of arrangements is 3! = 6. So total is 24 * 6 = 144. However, if we fix P's position, how does that compare? If we fix P's position, then the block can be arranged in 2 ways (R-P-T or T-P-R) and the rest can be arranged in 4! ways. Wait, but earlier we had 2 * 4! = 48. But according to this, it's 144. There's a discrepancy here. Which is correct?

Wait, if we don't fix positions, the total number of circular arrangements where all three are together is 144. However, if we fix P's position, then the number of such arrangements is 2 * 4! = 48. Since fixing P's position reduces the total number of arrangements by a factor of 7 (since there are 7 positions). Wait, no. Wait, fixing P's position is a way to account for circular rotations. So if the total number without fixing is 144, then fixing P's position would give 144 / 7? No, that doesn't make sense. Wait, actually, when we fix P's position, we are effectively dividing by 7 (the number of rotations). So if the total number of arrangements where all three are together is 144, then fixing P's position would give 144 / 7? But that's not an integer. Wait, something is wrong here.

Wait, let me recast. The total number of circular arrangements where all three are together is calculated as follows: treat the block as a single entity, so we have 5 entities. The number of circular arrangements is (5-1)! = 24. Multiply by the number of ways to arrange the block: 3! = 6. So total is 24 * 6 = 144. However, if we fix P's position, then how many such arrangements are there? If we fix P's position, then the block can be arranged in two ways: R-P-T or T-P-R, and the rest of the people can be arranged in 4! ways. So that's 2 * 4! = 48. But since fixing P's position accounts for circular rotations, the total number of arrangements without fixing is 48 * 7? No, that would be overcounting. Wait, no. If we fix P's position, we are considering only arrangements where P is in a specific seat, so the total number of arrangements where all three are together is 48. But according to the previous calculation, it's 144. Therefore, there's a contradiction. Which is correct?

Wait, let's think of a smaller case. Suppose we have 3 people: P, R, T. How many circular arrangements are there where all three are together? Well, since it's a circular table, there's only 1 arrangement (since rotations are considered the same). But according to the formula: treat them as a block, so 1 entity, circular arrangements: (1-1)! = 1, multiplied by 3! = 6. Wait, that gives 6, which is wrong. Wait, clearly, the formula is wrong here. Wait, no. Wait, if we have 3 people, the number of circular arrangements is (3-1)! = 2. But if we treat them as a block, we have 1 entity, so (1-1)! = 1, multiplied by 3! = 6. Which is wrong. Therefore, my previous approach is flawed.

Ah! I see the mistake. When we treat the block as a single entity, the formula for circular permutations is (number of entities - 1)! So if we have a block of 3 people and 4 other people, total entities is 5, so (5-1)! = 24. But if we have only the block, then it's (1-1)! = 1. However, in reality, when we have a block of 3 people, the number of circular arrangements is (number of entities - 1)! multiplied by the arrangements within the block. Wait, but in the case of only the block, it's (1-1)! * 3! = 1 * 6 = 6, but the actual number of circular arrangements for 3 people is 2. Therefore, there's a problem here. So my previous approach is wrong.

Wait, so what's the correct way to calculate the number of circular arrangements where all three are together? Let me think again. If 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: treat the k people as a single block, so we have (n - k + 1) entities. The number of circular arrangements is ( (n - k + 1) - 1 )! = (n - k)! multiplied by the number of ways to arrange the k people within the block, which is k!. Therefore, total is (n - k)! * k!.

Wait, let's test this with n=3, k=3. Then it would be (3 - 3)! * 3! = 0! * 6 = 1 * 6 = 6. But the actual number of circular arrangements for 3 people is 2. So this formula is also wrong. Therefore, my approach is incorrect.

Wait, maybe the formula is different. Let me recall. For linear arrangements, the number of ways to have a block of k people is (n - k + 1)! * k!. For circular arrangements, it's different. Let me think. When arranging in a circle, fixing one person's position to remove rotational symmetry. Suppose we fix one person in the block, say P. Then the rest of the block can be arranged in (k - 1)! ways, and the remaining (n - k) people can be arranged in (n - k)! ways. Therefore, total number of circular arrangements is (k - 1)! * (n - k)!.

Wait, let's test this with n=3, k=3. Then it would be (3 - 1)! * (3 - 3)! = 2! * 0! = 2 * 1 = 2, which is correct. For n=4, k=2: number of circular arrangements where two specific people are together. Fix one person, say P, then the other person can be in 1 seat adjacent to P (since it's a circular table with 4 seats). Wait, no. If we fix P's position, the other person can be in 2 seats adjacent to P. Wait, but if we want them to be together, then the other person must be in one of the two seats adjacent to P. Then the remaining 2 people can be arranged in 2! ways. So total arrangements: 2 * 2! = 4. According to the formula: (k - 1)! * (n - k)! = (2 - 1)! * (4 - 2)! = 1! * 2! = 1 * 2 = 2. Which is different. So this formula is also wrong.

Hmm, this is getting too confusing. Maybe I need to find another way. Let me go back to the original problem.

Earlier, I had two different approaches: one where I calculated the valid arrangements by arranging the other people first and placing P, R, T in the gaps, getting 144, and another approach using inclusion-exclusion which is getting tangled up. Maybe I should trust the first method?

Let me think again. If I fix one of the non-P, R, T people, say person A, then arrange the remaining 3 non-P, R, T people around the table. That would be 3! ways. Then, there are 4 gaps between the 4 people (including person A). We need to place P, R, T into these gaps, one per gap, so that they are not adjacent. The number of ways to choose the gaps is C(4, 3) = 4, and then arrange P, R, T in those gaps: 3! ways. So total arrangements: 3! * 4 * 3! = 6 * 4 * 6 = 144.

But let me check with a smaller case. Suppose there are 4 people: P, R, T, and one other person, say A. How many ways can they sit around a table so that P, R, T are not all adjacent? Wait, but with 4 people, the condition would be that no two of P, R, T are adjacent. Wait, but with 4 people, if we have 3 people who can't be adjacent, it's impossible because there are only 4 seats. Wait, maybe not. Let me think. If we have 4 seats, and 3 people P, R, T who can't be adjacent. Wait, but in a circular table with 4 seats, each person has two neighbors. If we have 3 people, they must occupy 3 seats, so at least two of them will be adjacent. Therefore, the number of valid arrangements is zero. Let me check with the first method. Fix person A. Then arrange the remaining 0 non-P, R, T people (since total non-P, R, T is 1, which is A). So arranging 0 people: 1 way. Then, number of gaps is 1 (since there's only person A). Wait, no. If we fix person A, then there are 3 seats left. Wait, no. Wait, total people are 4: A, P, R, T. If we fix A's position, then the remaining 3 people (P, R, T) need to be arranged around the table. But the condition is that no two of P, R, T are adjacent. However, with 3 people in 3 seats around the table, they will all be adjacent to each other. Therefore, there are zero valid arrangements. According to the first method: arranging the non-P, R, T people (which is just A) with fixing A's position, so 0! = 1 way. Then, number of gaps is 1 (since there's only A). We need to place 3 people into 1 gap, which is impossible. Therefore, the number of ways is zero, which matches. So that seems correct.

Another test case: 5 people, with P, R, T needing to not sit next to each other. Let's see. Total people: 5. Non-P, R, T: 2 people, say A and B. Using the first method: fix A's position. Arrange the remaining 1 non-P, R, T person (B) in 1! = 1 way. Then, number of gaps is 2 (between A and B, and between B and A). Wait, no. If we have 2 people (A and B) fixed around the table, there are 2 gaps. We need to place P, R, T into these gaps, one per gap. But there are 2 gaps and 3 people, which is impossible. Therefore, the number of valid arrangements is zero. But is that true? Let's see. In a circular table with 5 seats, can we seat P, R, T such that none are adjacent? Let's try. If we have A and B seated, then the seats are A, ?, ?, ?, B. Wait, no. If we fix A's position, then B can be seated in one of the remaining seats. Wait, with 5 seats, fixing A's position, B can be in seat 2, 3, 4, or 5. Wait, but if we have two people (A and B), then there are 5 - 2 = 3 seats left for P, R, T. Wait, no. Wait, total people are 5: A, B, P, R, T. If we fix A's position, then B can be seated in any of the remaining 4 seats. Wait, but in the first method, I considered arranging the non-P, R, T people (A and B). So if we fix A's position, then B can be arranged in 1! = 1 way? Wait, no. Wait, if there are 2 non-P, R, T people (A and B), and we fix A's position, then B can be arranged in (2 - 1)! = 1! = 1 way? Wait, no. Wait, the formula for arranging n people around a table is (n-1)! So if we have 2 non-P, R, T people, fixing A's position, then B can be arranged in 1! = 1 way. So the two non-P, R, T people are A and B, fixed with A in one seat and B in another. Then, the number of gaps between them is 2. So we need to place P, R, T into these gaps. But there are 2 gaps and 3 people, which is impossible. Therefore, the number of valid arrangements is zero. But is that true? Let's see. If we have 5 seats, and we need to seat P, R, T such that none are adjacent. Let's try to place them. Suppose we have A and B seated with one seat between them. Then, the seats are A, X, B, Y, Z. Wait, but with 5 seats, if A and B are seated with one seat between them, then there are two seats between A and B. Wait, this is getting too confusing. Maybe it's better to accept that the first method works for the test case where it's impossible, giving zero, which is correct. So maybe the first method is correct.

Therefore, returning to the original problem, with 7 people, the first method gives 144. However, let me check with inclusion-exclusion. Earlier, I had:

Total arrangements: 720.

|A ∪ B ∪ C| = 3*240 - 3*48 + 144 = ?

Wait, earlier I had |A| = 240, |B| = 240, |C| = 240. Then |A ∩ B| = |A ∩ C| = |B ∩ C| = 48 each. And |A ∩ B ∩ C| = 144. Wait, but earlier I was confused about |A ∩ B ∩ C|. If I use the formula from the first method, where the number of arrangements where all three are together is 144, then:

|A ∪ B ∪ C| = 3*240 - 3*48 + 144 = 720 - 144 + 144 = 720? That can't be. Wait, no:

Wait, inclusion-exclusion formula is |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.

So plugging in:

= 3*240 - 3*48 + 144 = 720 - 144 + 144 = 720. Which would mean that |A ∪ B ∪ C| = 720, so the number of arrangements with no two of P, R, T together is total - |A ∪ B ∪ C| = 720 - 720 = 0, which is impossible. Clearly, something is wrong here.

This suggests that my calculation of |A ∩ B ∩ C| is wrong. If |A ∪ B ∪ C| = 720, then it would mean that all arrangements have at least two of P, R, T together, which is not true. For example, the first method suggests there are 144 valid arrangements. Therefore, my calculation of |A ∩ B ∩ C| is incorrect.

Let me try again. Let's recast the inclusion-exclusion.

First, |A|, |B|, |C|: each is the number of arrangements where a specific pair is together. Earlier, I thought |A| = 240. Let me verify this again.

Total number of circular arrangements: 720.

Number of arrangements where P and R are together: ?

As before, fixing P's position, R can be in 2 seats adjacent to P, and the rest can be arranged in 5! ways. Wait, no. If we fix P's position, then there are 6 seats left. R can be in 2 seats adjacent to P. Then the remaining 5 people can be arranged in 5! ways. So total arrangements with P and R together is 2 * 5! = 240. Since fixing P's position accounts for circular rotations, the total number is 240. Therefore, |A| = 240. Similarly for |B| and |C|.

Now, |A ∩ B|: arrangements where P and R are together AND P and T are together. As discussed earlier, this is equivalent to arrangements where P is adjacent to both R and T. How many are there?

If we fix P's position, then R and T must be in the two seats adjacent to P. The number of ways is 2! (arrangements of R and T) * 5! (arrangements of the rest)? Wait, no. If we fix P's position, then R and T are in the two seats adjacent to P, and the remaining 4 people can be arranged in 4! ways. Wait, total seats are 7. Fix P's position, then there are 6 seats left. If R and T are in two of those seats (adjacent to P), then the remaining 4 seats are for the other 4 people. So the number of arrangements is 2! * 4! = 2 * 24 = 48. Therefore, |A ∩ B| = 48. Similarly for |A ∩ C| and |B ∩ C|.

Now, |A ∩ B ∩ C|: arrangements where P and R are together, P and T are together, and R and T are together. This is the case where all three are together. How many such arrangements are there?

If we fix P's position, then R and T must be adjacent to P and also adjacent to each other. So the three of them form a block. As discussed earlier, there are two possible blocks: R-P-T or T-P-R. Then, the remaining 4 people can be arranged in 4! ways. So total arrangements with P fixed is 2 * 4! = 48. Therefore, the total number of such arrangements without fixing is 48 * 7 / 7? Wait, no. Wait, fixing P's position accounts for circular rotations, so the total number of arrangements where all three are together is 48. But earlier, when I thought of treating them as a block, I had a different number. Let me check with another approach.

If we treat P, R, T as a single block, then we have 5 entities: the block and the other 4 people. The number of circular arrangements is (5-1)! = 24. Multiply by the number of ways to arrange the block: 3! = 6. So total is 24 * 6 = 144. But according to fixing P's position, it's 48. Which is correct?

Wait, if we don't fix positions, then the total number of circular arrangements where all three are together is 144. But if we fix P's position, it's 48. Since fixing P's position reduces the total number by a factor of 7 (number of seats), then 144 / 7 is not an integer, which is impossible. Therefore, there must be an error in one of the calculations.

Wait, let's think of it this way: when we fix P's position, the number of arrangements where all three are together is equal to the number of ways to arrange R and T next to P and also adjacent to each other. If P is fixed, then R and T must be on either side of P, forming a block of R-P-T or T-P-R. Then, the rest of the people can be arranged in 4! ways. So that's 2 * 4! = 48. Therefore, the total number of such arrangements without fixing is 48 * 7 / 7? No, fixing P's position already accounts for all rotations. Therefore, the total number of arrangements where all three are together is 48. But according to the block method, it's 144. There's a contradiction here.

Wait, let's think of the total number of circular arrangements where all three are together. If we have 7 people, and we want to count the number of circular arrangements where P, R, T are all together.

First, consider the number of ways to arrange the block of P, R, T and the other 4 people. In circular arrangements, the number is (number of entities - 1)! * arrangements within the block. Number of entities is 5 (block + 4 people). So (5-1)! = 24. Multiply by 3! = 6 for the block arrangements. Total is 24 * 6 = 144.

But if we fix P's position, then the number of such arrangements is 2 * 4! = 48. Since fixing P's position accounts for all rotations, the total number should be 48. But according to the block method, it's 144. This suggests that the block method is wrong.

Wait, but if we fix P's position, then the number of arrangements where all three are together is 48. Therefore, the total number of such arrangements without fixing is 48 * 7 / 7 = 48? No, that doesn't make sense. Wait, fixing P's position is just one specific arrangement. The total number of circular arrangements is 720. If we fix P's position, we're considering 720 / 7 = 102.857... which is not an integer. Wait, no. Wait, fixing P's position is a way to count the number of arrangements where P is in a specific seat. The total number of circular arrangements is (7-1)! = 720. If we fix P's position, the number of arrangements is 6! = 720. Wait, no! Wait, fixing P's position in a circular table reduces the problem to arranging the remaining 6 people linearly, which is 6! = 720. Wait, but that can't be, because the total number of circular arrangements is 720. So fixing P's position doesn't change the count; it's just a way to count them.

Wait, I think I'm getting confused between fixing a position and counting. Let me clarify:

Total number of circular arrangements: (7-1)! = 720.

If we fix P's position, then the number of arrangements is 6! = 720. Wait, that can't be. Wait, no. If you fix one person's position in a circular table, the number of arrangements is (n-1)! So for n=7, fixing P's position, the number of arrangements is 6! = 720? No, that's not possible. Wait, no. Wait, fixing one person's position in a circular table, the number of arrangements is (n-1)! So for n=7, it's 6! = 720. But the total number of circular arrangements is also 6! = 720. So fixing P's position is just a way to count them.

Therefore, if we fix P's position, and calculate the number of arrangements where all three are together, it's 2 * 4! = 48. Therefore, the total number of such arrangements without fixing is 48. But according to the block method, it's 144. This suggests that the block method is wrong.

Wait, but if we fix P's position, and calculate the number of arrangements where all three are together as 48, then the total number of such arrangements is 48. But if we don't fix P's position, then it should be 48 * 7 / 7 = 48? No, that doesn't make sense. Wait, no. If we fix P's position, we are considering all arrangements where P is in a specific seat. The total number of arrangements where all three are together is equal to the number of such arrangements with P fixed multiplied by 7 (number of seats P can be in)? No, because fixing P's position already accounts for all rotations.

This is very confusing. Maybe I need to find another way. Let me think of the number of ways where all three are together as follows:

First, choose the position of the block. In a circular table, the number of ways to place a block of 3 people is 7 (since it can start at any of the 7 seats). However, since it's a circular table, each block is counted 3 times (once for each seat in the block). Wait, no. For a block of 3 people, there are 7 possible starting positions, but each block is counted 3 times (once for each person in the block). Therefore, the number of distinct blocks is 7 / 3, which is not an integer. This suggests that my approach is wrong.

Alternatively, the number of ways to arrange the block is 7 (positions for the block) * 3! (arrangements within the block) * 4! (arrangements of the rest). But since it's a circular table, we need to divide by 7 to account for rotations. Wait, this is getting too tangled.

Maybe I should accept that my inclusion-exclusion approach is leading to contradictions and trust the first method which gave 144. Let me check with another approach.

The first method: arranging the 4 non-P, R, T people around the table, then placing P, R, T in the gaps.

Number of ways to arrange the 4 people: (4-1)! = 6.

Number of gaps: 4.

Choose 3 gaps: C(4,3) = 4.

Arrange P, R, T in the gaps: 3! = 6.

Total: 6 * 4 * 6 = 144.

This seems logical. Let me think of it as arranging the 4 people first, which creates 4 gaps. Placing P, R, T in separate gaps ensures they are not adjacent.

Alternatively, if I use the inclusion-exclusion, and assume that |A ∩ B ∩ C| is 48, then:

|A ∪ B ∪ C| = 3*240 - 3*48 + 48 = 720 - 144 + 48 = 624.

Then, the number of valid arrangements would be 720 - 624 = 96.

But this contradicts the first method's result of 144.

Alternatively, if |A ∩ B ∩ C| is 144, then:

|A ∪ B ∪ C| = 720 - 144 + 144 = 720, which is impossible.

Alternatively, if |A ∩ B ∩ C| is 24, then:

|A ∪ B ∪ C| = 720 - 144 + 24 = 504, and valid arrangements would be 720 - 504 = 216.

But this is just guessing.

Alternatively, let's think of |A ∩ B ∩ C| as the number of arrangements where all three are together, which is 144. But if we use the first method, the valid arrangements are 144, which would mean that the number of invalid arrangements is 720 - 144 = 576. But according to inclusion-exclusion, if |A ∪ B ∪ C| = 576, then:

576 = 3*240 - 3*|A ∩ B| + |A ∩ B ∩ C|.

So 576 = 720 - 3*|A ∩ B| + |A ∩ B ∩ C|.

Rearranged: 3*|A ∩ B| - |A ∩ B ∩ C| = 720 - 576 = 144.

If we assume |A ∩ B| = 48, then 3*48 - |A ∩ B ∩ C| = 144 => 144 - |A ∩ B ∩ C| = 144 => |A ∩ B ∩ C| = 0, which is impossible.

If we assume |A ∩ B| = 48 and |A ∩ B ∩ C| = 144, then 3*48 - 144 = 144 - 144 = 0 ≠ 144.

This is not working.

Maybe my calculation of |A ∩ B| is wrong. Let me think again.

If |A ∩ B| is the number of arrangements where P is adjacent to both R and T, which includes both cases where all three are together and where P is between R and T but R and T are not adjacent.

If I calculate this as follows:

Number of ways where P is adjacent to both R and T = number of ways where P is between R and T (R-P-T or T-P-R) + number of ways where all three are together.

Wait, but if all three are together, then P is between R and T. So actually, the number of arrangements where P is adjacent to both R and T is equal to the number of arrangements where all three are together. Wait, no. If all three are together, then P is adjacent to both R and T, but also R and T are adjacent. If P is between R and T but they are not together, then P is adjacent to both R and T, but R and T are not adjacent. So the total number of arrangements where P is adjacent to both R and T is equal to the number of arrangements where all three are together plus the number of arrangements where P is between R and T but R and T are not adjacent.

But how do we calculate this?

If we fix P's position, then the number of ways where P is adjacent to both R and T is 2 * 4! = 48, as before. This includes both cases where R and T are adjacent (all three together) and where they are not.

But how many of these are all three together? If we fix P's position, the number of arrangements where all three are together is 2 * 4! = 48? No, earlier we thought it was 2 * 4! = 48, but that seems to include both cases. Wait, no. If we fix P's position, and want all three to be together, then the block can be in two positions (to the left or right of P), and within the block, R and T can be arranged in 2 ways. So total is 2 * 2 * 4! = 96? No, if we fix P's position, and want all three together, then the block can be in two positions (positions 1-2-3 or 1-7-6), and within each block, R and T can be arranged in 2 ways. So total is 2 * 2 * 4! = 96? But this would mean that the number of arrangements where all three are together is 96, and the number of arrangements where P is adjacent to both R and T but they are not together is 48 - 96? No, this is getting too tangled.

At this point, I think I need to accept that my inclusion-exclusion approach is error-prone due to the complexity of overlapping cases, and trust the first method which gave 144.

But to verify, let's think of another way. The total number of ways where no two of P, R, T are adjacent is equal to the total number of circular permutations minus the number of permutations where at least two are adjacent.

But if the first method gives 144, and the total is 720, then the number of invalid arrangements is 720 - 144 = 576.

Alternatively, let's calculate it using the first method's logic.

Number of ways to arrange the 4 non-P, R, T people: (4-1)! = 6.

Number of ways to place P, R, T in the gaps: C(4,3) * 3! = 4 * 6 = 24.

Total: 6 * 24 = 144.

This seems logical.

Alternatively, if we use the formula for non-adjacent arrangements in circular tables.

For linear arrangements, the number of ways to arrange n people with k people not adjacent is C(n - k + 1, k) * k! * (n - k)! but for circular tables, it's different.

However, I recall that for circular arrangements, the formula is similar to linear but adjusted for circularity.

In this case, we have 7 seats. We want to place 3 people (P, R, T) such that no two are adjacent, and the rest are 4 people.

The formula for the number of ways to choose seats for P, R, T is C(n - k, k) + C(n - k - 1, k - 1) for circular tables? No, I need to recall.

Actually, the number of ways to choose k non-consecutive seats around a circular table of n seats is C(n - k, k) + C(n - k - 1, k - 1). But I'm not sure.

Alternatively, for circular tables, the number of ways to choose k seats such that no two are adjacent is C(n - k, k) + C(n - k - 1, k - 1).

But let me check with n=7, k=3.

C(7 - 3, 3) + C(7 - 3 - 1, 3 - 1) = C(4,3) + C(3,2) = 4 + 3 = 7.

But the number of ways to choose 3 non-consecutive seats around a circular table of 7 seats is 7? Let me think.

In a circular table with 7 seats, how many ways to choose 3 seats with no two adjacent?

This is equivalent to placing 3 people and 4 non-people around the table, no two people adjacent.

This is similar to the problem of arranging objects with spacing.

For circular arrangements, the formula is C(n - k, k) + C(n - k - 1, k - 1).

But if n=7, k=3, it gives 4 + 3 = 7. Let me verify.

Imagine the 7 seats. To choose 3 seats with no two adjacent.

This is equivalent to placing 3 people and 4 non-people around the table, with at least one non-person between each pair of people.

In circular arrangements, this is equal to (n - k) choose k divided by something?

Actually, the formula for the number of ways to choose k non-consecutive seats around a circular table of n seats is (n / (n - k)) * C(n - k, k). Wait, no.

Alternatively, it's C(n - k, k) + C(n - k - 1, k - 1). For n=7, k=3, it's C(4,3) + C(3,2) = 4 + 3 = 7.

Let me count manually.

In a circular table with 7 seats, how many ways to choose 3 seats with no two adjacent?

Let's fix one seat as seat 1.

If seat 1 is chosen, then seats 2 and 7 cannot be chosen. We need to choose 2 more seats from seats 3, 4, 5, 6, with no two adjacent.

This is equivalent to choosing 2 non-consecutive seats from 4 seats in a line (since seat 1 is fixed, and seats 2 and 7 are excluded). The number of ways is C(4 - 2 + 1, 2) = C(3, 2) = 3.

If seat 1 is not chosen, then we need to choose 3 seats from seats 2-7, with no two adjacent. This is equivalent to choosing 3 non-consecutive seats in a line of 6 seats, which is C(6 - 3 + 1, 3) = C(4, 3) = 4.

Wait, but this is for linear arrangements. For circular, if we fix seat 1 as not chosen, it becomes a linear arrangement. So total number is 3 + 4 = 7, which matches the formula.

Therefore, the number of ways to choose the seats is 7.

Once the seats are chosen, the number of ways to arrange P, R, T is 3! = 6.

The number of ways to arrange the other 4 people is (4-1)! = 6 (since it's a circular table). Wait, no. Wait, if we have already fixed the seats for P, R, T, then the remaining 4 people can be arranged in 4! ways, but since the table is circular, we need to adjust for rotations. Wait, no. Wait, if we have already fixed the positions of P, R, T, then the remaining 4 people can be arranged in 4! ways, but since the table is circular, but we have already fixed the positions of P, R, T, which might have fixed the rotation.

Wait, this is getting complicated. Let me think.

If we first choose the seats for P, R, T, which are 3 specific seats with no two adjacent. Once these seats are chosen, the rest of the seats are for the other 4 people. Since the table is circular, but we have already fixed the positions of P, R, T, the remaining people can be arranged in 4! ways. However, since the table is circular, but we have already fixed the positions of P, R, T, which might have fixed the rotation.

Wait, no. If we choose the seats for P, R, T, then the arrangement is determined up to rotation. But since we have already chosen specific seats, the rotations are already accounted for.

Wait, this is confusing. Let me think of it as follows:

The total number of ways to arrange the 7 people with P, R, T not adjacent is equal to the number of ways to choose the seats for P, R, T (7 ways) multiplied by the number of ways to arrange P, R, T in those seats (3!) multiplied by the number of ways to arrange the other 4 people in the remaining seats (4!).

But since it's a circular table, we need to divide by the number of rotations? No, because we have already fixed the seats by choosing specific positions.

Wait, no. If we choose specific seats, then the number of arrangements is 7 (choices of seats) * 3! * 4! = 7 * 6 * 24 = 1008. But this is greater than the total number of circular arrangements (720), which is impossible.

Clearly, this approach is wrong.

The error is that choosing specific seats in a circular table already accounts for rotations. Wait, no. If we count the number of ways to choose seats as 7, and then multiply by arrangements, we are overcounting.

Actually, the correct way is:

The number of ways to arrange the 7 people with P, R, T not adjacent is equal to (number of ways to choose seats for P, R, T) * (number of ways to arrange P, R, T) * (number of ways to arrange the others).

But in circular arrangements, the number of ways to choose seats for P, R, T is 7 (as calculated earlier). However, this 7 already includes rotations. Wait, no. If we fix the table, the number of ways to choose 3 non-adjacent seats is 7, as calculated. Then, for each such choice, we can arrange P, R, T in 3! ways, and the others in 4! ways. But since the table is circular, we need to divide by 7 to account for rotations? No, because we have already fixed the seats.

Wait, this is very confusing. Let me think of it as linear arrangements.

If it were a linear table, the number of ways would be:

Number of ways to choose seats for P, R, T: C(n - k + 1, k) = C(7 - 3 + 1, 3) = C(5, 3) = 10.

Then, arrange P, R, T: 3! = 6.

Arrange the others: 4! = 24.

Total: 10 * 6 * 24 = 1440.

But for a circular table, it's different.

However, in our first method, we got 144, which is 1440 / 10 = 144. So perhaps for circular tables, we need to divide by n?

Wait, if for linear it's 1440, and for circular it's 1440 / 7 = 205.714... which is not an integer.

This suggests that this approach is also incorrect.

Given the time I've spent and the fact that the first method gave a logical result that passed the test case, I think I should go with the first method's answer of 144.

Therefore, the number of ways is 144.

But let me check with another approach.

The formula for the number of ways to arrange n people around a circular table with k people not adjacent is:

(n - k)! * C(n - k, k) + ...? No, I need to recall.

Alternatively, using the principle of inclusion-exclusion, but I need to resolve the earlier contradiction.

Let me try again with the inclusion-exclusion, assuming that |A ∩ B ∩ C| is 48.

So, |A ∪ B ∪ C| = 3*240 - 3*48 + 48 = 720 - 144 + 48 = 624.

Then, the number of valid arrangements is 720 - 624 = 96.

But this contradicts the first method's 144.

Alternatively, if |A ∩ B ∩ C| is 24, then:

|A ∪ B ∪ C| = 720 - 144 + 24 = 504, and valid arrangements = 720 - 504 = 216.

But how to determine |A ∩ B ∩ C|?

If we use the first method's result of 144 valid arrangements, then the invalid arrangements are 720 - 144 = 576.

So, 576 = |A ∪ B ∪ C| = 3*240 - 3*|A ∩ B| + |A ∩ B ∩ C|.

So, 576 = 720 - 3*|A ∩ B| + |A ∩ B ∩ C|.

Rearranged: 3*|A ∩ B| - |A ∩ B ∩ C| = 720 - 576 = 144.

If we assume that |A ∩ B| = 48, then 3*48 - |A ∩ B ∩ C| = 144 => 144 - |A ∩ B ∩ C| = 144 => |A ∩ B ∩ C| = 0, which is impossible.

If we assume |A ∩ B| = 56, then 3*56 - |A ∩ B ∩ C| = 144 => 168 - |A ∩ B ∩ C| = 144 => |A ∩ B ∩ C| = 24.

But how to find |A ∩ B|?

If |A ∩ B| is the number of arrangements where P is adjacent to both R and T, which we calculated as 48 when fixing P's position. If we don't fix P's position, it would be 48 * 7 / 7 = 48? No, this is not clear.

Alternatively, if |A ∩ B| is 48, and |A ∩ B ∩ C| is 24, then the equation holds.

But how to verify |A ∩ B ∩ C| = 24?

If we fix P's position, the number of arrangements where all three are together is 2 * 4! = 48. But if we don't fix P's position, it would be 48 * 7 / 7 = 48? No, this is not correct.

Alternatively, if the number of arrangements where all three are together is 144, then |A ∩ B ∩ C| = 144. Plugging into the equation: 3*|A ∩ B| - 144 = 144 => 3*|A ∩ B| = 288 => |A ∩ B| = 96.

But earlier, we calculated |A ∩ B| as 48.

This is very confusing.

Given the time I've spent and the fact that the first method gave a logical result, I think I should conclude that the correct answer is 144, and the inclusion-exclusion approach has errors in calculating the intersections.

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

**Final Answer**
\boxed{144}

To solve the problem of determining how many 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 proceed with a structured approach that avoids overcounting due to rotational symmetry and ensures the adjacency condition is respected.

---

### Step 1: Total Circular Permutations

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

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

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

---

### Step 2: Strategy to Avoid Adjacency

We want to **avoid** any arrangement where **P, R, and T are adjacent to each other**. A good strategy is to **first arrange the other 4 people** (let's call them A, B, C, D), and then **place P, R, T in the gaps** between them such that no two of them are adjacent.

This ensures that P, R, and T are **not adjacent** to each other.

---

### Step 3: Arrange the 4 Non-Adjacent People

We fix one of the 4 people (say, A) to eliminate rotational symmetry. The remaining 3 people (B, C, D) can be arranged in:

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

This gives us 6 distinct arrangements of the 4 people around the table.

---

### Step 4: Count the Gaps for P, R, T

When 4 people are seated around a circular table, they create **4 gaps** between them (one between each pair of adjacent people). These are the positions where we can place P, R, and T such that they are **not adjacent** to each other.

We need to choose **3 out of these 4 gaps** to place P, R, and T. The number of ways to choose 3 gaps from 4 is:

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

Then, we can arrange P, R, and T in these 3 chosen gaps in:

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

---

### Step 5: Total Valid Arrangements

Now, multiply the number of ways to arrange the 4 people, the number of ways to choose the gaps, and the number of ways to arrange P, R, and T in those gaps:

$$
\text{Total valid arrangements} = 3! \times \binom{4}{3} \times 3! = 6 \times 4 \times 6 = 144
$$

---

### 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.