Explanation:
Okay, so I need to figure out how many ways 8 people can sit around a round table, but with the condition that Pierre, Rosa, and Thomas all want to sit together. Hmm, circular permutations can be a bit tricky because rotations are considered the same. Let me think through this step by step.

First, I remember that for circular permutations of n people, the number of distinct arrangements is (n-1)! instead of n! because rotations are considered the same. But here, there's a condition: three specific people must sit together. So I need to adjust for that.

Let me recall, when dealing with groups that must stay together in circular permutations, we can treat the group as a single entity or "block." So if Pierre, Rosa, and Thomas must sit together, maybe I can consider them as one block and then arrange the blocks around the table. But since it's a circular table, I need to be careful with how I count the arrangements.

So, normally, if we have 8 people, and we're grouping 3 of them into a block, we're effectively creating (8 - 3 + 1) = 6 "items" to arrange around the table: the block and the other 5 individuals. But wait, actually, the total number of items would be 1 block + 5 other people = 6 items. However, in circular permutations, the number of ways to arrange n items is (n-1)! So here, arranging 6 items around a circular table would be (6-1)! = 5!.

But then, within the block, the three people can be arranged among themselves in 3! ways. So total arrangements would be 5! * 3!.

But let me check if that's correct. Let me think again. If it's a circular table, fixing the block's position to account for rotations? Wait, but when we treat the block as a single entity, do we still need to adjust for rotations? Hmm.

Alternatively, sometimes people use the formula for circular permutations with a group: fix the position of the block to eliminate rotational symmetry. Let me think. For example, if you have a block of people, you can fix the block in a certain position, then arrange the rest relative to it. But since rotations are considered the same, fixing the block's position would account for that.

But if I consider the block as a single entity, then arranging the block and the other 5 people around the table is (number of entities - 1)! So (6 - 1)! = 5! as before. Then multiply by the number of internal arrangements of the block, which is 3!.

So that would give 5! * 3! = 120 * 6 = 720. Wait, but let me check if that's correct.

But wait, let me think again. Suppose there are no restrictions. The total number of circular permutations is 7! / 8 = 5040 / 8 = 630? Wait, no. Wait, circular permutations formula is (n-1)! So for 8 people, it's 7! = 5040. Wait, no! Wait, no. Wait, hold on. Wait, (n-1)! is the formula. So for 8 people, it's 7! = 5040. But if we have a restriction that three people must sit together, then the method I mentioned before would be treating them as a block, so 6 entities, so (6-1)! = 5! arrangements, multiplied by 3! for the block. So 5! * 3! = 120 * 6 = 720. But 720 is less than 5040, which seems plausible.

Wait, but is there another way to think about it?

Alternatively, suppose we fix one person's position to remove rotational symmetry. Let me try that approach. Let me fix Pierre's position. Since rotations are considered the same, we can fix Pierre at a specific seat. Then, we need to arrange the remaining 7 people around the table. But since Pierre, Rosa, and Thomas must all sit together, Rosa and Thomas have to be adjacent to Pierre or to each other?

Wait, if Pierre is fixed, then the block of Pierre, Rosa, and Thomas can be arranged in different ways. Let me think. If we fix Pierre's position, then the block can be considered as Pierre sitting in a specific seat, and Rosa and Thomas sitting next to him. However, since the block can be arranged in different orders.

But since we fixed Pierre's position, the block can be arranged in two directions: for example, Pierre, Rosa, Thomas or Pierre, Thomas, Rosa. So, the block can be arranged in 2! ways? Wait, no, since within the block, there are 3 people, so once Pierre is fixed, Rosa and Thomas can be arranged in 2! ways? Wait, but if the entire block is considered, maybe we need to consider the block's position.

Wait, if Pierre is fixed, then the block can be placed in different positions. Wait, but since the table is circular and we fixed Pierre's position, the block can be arranged either clockwise or counter-clockwise around Pierre?

Wait, this seems confusing. Let me think again. If we fix Pierre's position, then the remaining people are Rosa, Thomas, and 5 others. The condition is that Rosa and Thomas must be adjacent to Pierre or to each other? Wait, no, the condition is that all three must sit together. So the block of Pierre, Rosa, and Thomas must be together. Since Pierre is fixed, the block can be arranged in two directions: either going clockwise or counter-clockwise from Pierre.

Wait, but actually, if Pierre is fixed, then the block can be thought of as a sequence of three people: Pierre, then Rosa and Thomas next to him. But since there's three people in the block, if Pierre is fixed, the block can be arranged in two different ways? Or perhaps more?

Alternatively, fixing Pierre's position, the block of three can be arranged in 2! ways? Wait, no. Let me think. If we fix Pierre in a seat, then we can arrange Rosa and Thomas in the seats adjacent to Pierre. But since it's a block of three people, they need to occupy three consecutive seats. If Pierre is fixed, then the block can be placed either to his left or to his right? Wait, but in a circular table, if Pierre is fixed, then the seats are ordered clockwise. So if we fix Pierre at seat 1, then the block could occupy seats 1, 2, 3 or seats 1, 8, 7? Wait, but if we fix Pierre at seat 1, then the block must include seat 1. So the block can be seats 1, 2, 3 or seats 1, 8, 7. But if we consider that the block must be three consecutive seats, then fixing Pierre at seat 1, the block can be either seats 1, 2, 3 or 1, 8, 7. However, if we consider the entire block as three consecutive seats, then once Pierre is fixed, the block can be placed in two different linear arrangements? Wait, but actually, the three people can be arranged in different permutations.

Wait, perhaps this approach is overcomplicating. Let me check the standard approach for circular permutations with a block.

In linear permutations, if we have a block, we treat it as a single entity, so the number of arrangements is (n - k + 1)! * k! where n is total people and k is the size of the block. But for circular permutations, it's different because of rotational symmetry.

But here's a method: For circular permutations with a block, you can fix the position of the block to remove rotational symmetry. So if you have a block of size k, you can fix the block in a specific position, then arrange the remaining people. However, fixing the block's position would mean that the number of arrangements is (total number of items - 1)! * internal arrangements.

Wait, maybe I should think of the block as an entity. So total entities: 1 block + 5 other people = 6 entities. In circular permutations, arranging n entities around a circular table is (n-1)! So here, (6-1)! = 5! Then, multiply by the number of ways to arrange the people within the block, which is 3!.

Therefore, total number of arrangements is 5! * 3! = 120 * 6 = 720. So that would be the answer. But let me check if this is correct.

Alternatively, another approach: total number of circular arrangements where the three people are together is equal to the number of linear arrangements where the three are together, divided by something? Wait, no, because circular vs linear is different.

Alternatively, suppose we consider the three people as a single entity. Then, in circular permutations, the number of ways to arrange the entities is (number of entities - 1)! So that's (6 - 1)! = 5! Then multiply by the permutations within the block: 3!. Therefore, 5! * 3! = 720. So that seems to be the same as before.

But let me check with an example. Let's say there are fewer people. Suppose there are 3 people, and they all must sit together. Then, how many arrangements? Well, since it's a circular table, there's only 1 arrangement. Using the formula: number of entities is 1 (the block), so (1 - 1)! = 0! = 1, multiplied by 3! = 6. Wait, that gives 6, which is wrong. Wait, that can't be. So clearly, this formula isn't correct. So my previous reasoning is flawed.

Wait, what's the problem here? If there are 3 people sitting around a round table, and they must all sit together, there's actually only 1 way, because rotations are considered the same. However, according to the formula (number of entities - 1)! * internal permutations, which would be (1 - 1)! * 3! = 1 * 6 = 6. Which is wrong. Therefore, my formula is incorrect.

So there's a mistake in my reasoning. Let me figure out why.

Ah, so when the entire group is the block, then treating it as a single entity, but in circular permutations, when you have one entity, the number of arrangements is (1 - 1)! = 1, which is correct, but then multiplied by 3! gives 6, which is wrong. So clearly, there's an error here.

Wait, maybe when we have only one block, the number of circular arrangements is 1, as there's only one way to arrange a single block around a table. But if you multiply by the internal permutations, then 1 * 3! = 6. But in reality, the number of circular arrangements where all 3 people are sitting together is 1 (since rotations are considered the same), but if you consider rotations as different, then it would be 2 (since in a circular table, there are (n-1)! arrangements, so for n=3, it's 2. Wait, hold on. Let me clarify:

Wait, if we have 3 people around a circular table, the number of distinct seating arrangements is 2, not 1. Wait, no. Wait, for n people, circular permutations are (n-1)! So for 3 people, it's 2. However, if we consider that all three must sit together, but they are already seated together. So the number of arrangements is 2. However, according to the formula, treating the three as a single block, the number of arrangements is (number of blocks -1)! * internal permutations. Which would be (1 - 1)! * 3! = 1 * 6 = 6. Which is wrong.

So that approach is wrong. Therefore, my initial reasoning was incorrect. So why is that?

Ah, maybe because when we consider the block, if we have only one block, then the number of arrangements is 1 (for the block) but the internal permutations are 3! but in reality, fixing the block's position, the internal permutations are 3! but since rotations are already accounted for by fixing the block's position. Wait, but how?

Alternatively, maybe the formula is different. Maybe when we have a single block, the number of circular arrangements is (number of entities - 1)! * internal permutations. But if there is only one entity, then (1 - 1)! * 3! = 1 * 6 = 6. However, in reality, the number is 2. So clearly something is wrong here.

Therefore, there's a mistake in my approach. Let me think again. What's the correct way to count circular permutations with a block?

Let me refer back to standard methods. In circular permutations, when we have a group that must stay together, we can use the following approach:

First, consider the group as a single "super person". Then, the number of ways to arrange the "super person" and the other individuals around the table is (number of entities - 1)!.

However, this is only if we consider that the "super person" can be rotated. But actually, when we treat the group as a single entity, the number of circular arrangements is (number of entities - 1)!.

However, then we have to multiply by the number of ways to arrange the people within the group.

But in the case where the group is the entire set, this would lead to (1 - 1)! * 3! = 1 * 6 = 6. But for n=3, the correct number is 2. So why is this discrepancy?

Ah! Because in the case where the entire group is the block, we have to consider that in circular permutations, when we fix the block's position, the internal arrangements are counted as permutations. However, if we don't fix the block's position, then the formula is different. Wait, actually, I think the formula is correct when there are multiple blocks, but perhaps not when there's only one block?

Wait, no. Wait, let me think again. Suppose we have n people, and we want to arrange them around a circular table with a specific block. Let's say we have a block of size k. So total entities: m = n - k + 1. Then, arranging m entities around a circular table is (m - 1)! and multiplying by k! for internal arrangements. However, in the case where m = 1 (i.e., n = k), this gives (1 - 1)! * k! = 1 * k! = k! which is wrong because for circular permutations it's (n - 1)!.

Wait, so there's a contradiction here. So clearly, the formula is not applicable when the entire set is a block. Therefore, my previous reasoning is wrong. Therefore, I need another approach.

So what is the correct way?

Let me think. Let me recall that when dealing with circular permutations with groups, the formula is similar to linear permutations but adjusted for circularity. For linear permutations, the number would be (n - k + 1)! * k! So for example, for 3 people in a line, it's (3 - 3 + 1)! * 3! = 1! * 6 = 6, which is correct. But for circular permutations, perhaps it's different.

Wait, another approach: For circular permutations, we can fix one person's position to remove rotational symmetry. Let me try that.

Let me fix Pierre's position. Since rotations are considered the same, fixing Pierre at a specific seat. Then, the problem reduces to arranging the remaining 7 people around the table with the condition that Rosa and Thomas are adjacent to Pierre or to each other? Wait, no, the condition is that all three (Pierre, Rosa, Thomas) must sit together. Since Pierre is fixed, the block of three must include Pierre and the two others. So, the block is Pierre, Rosa, and Thomas, and they need to be consecutive.

Since Pierre is fixed, the block can be arranged in two directions: either clockwise or counter-clockwise from Pierre. Wait, but how does that work?

Let me visualize the table. Let's fix Pierre at seat 1. Then, the block of three can occupy seats 1, 2, 3 or seats 1, 8, 7. These are the only two possible blocks that include seat 1 and are consecutive. Therefore, there are two possible positions for the block.

But within each block, the three people can be arranged in different orders. However, since Pierre is fixed at seat 1, the block's arrangement is determined by the positions of Rosa and Thomas.

Wait, for example, if the block occupies seats 1, 2, 3, then seat 1 is Pierre, so seats 2 and 3 must be Rosa and Thomas in some order. Similarly, if the block occupies seats 1, 8, 7, then seat 1 is Pierre, seat 8 and seat 7 must be Rosa and Thomas in some order. Therefore, for each block position (two possibilities), we have 2! arrangements for Rosa and Thomas. Therefore, total arrangements for the block is 2 (positions) * 2! (internal arrangements). Wait, but Pierre is fixed, so is that correct?

Wait, if the block can be in two positions, and in each position, the other two people can be arranged in 2! ways, so total of 2 * 2! = 4 ways. Then, once the block is placed, the remaining 5 people can be arranged in the remaining 5 seats in 5! ways. Therefore, total number of arrangements would be 4 * 5!.

But let me check that. Wait, if we fix Pierre's seat, then the block can be in two positions (left or right side). For each position, Rosa and Thomas can switch places, so 2! ways. Then, the remaining 5 people can be arranged in the remaining 5 seats. Therefore, total is 2 * 2! * 5!.

But then 2 * 2! * 5! = 2 * 2 * 120 = 480. But let me check with the previous formula (which gave 720). Which one is correct?

Alternatively, let me check the case where n=3. If there are 3 people, and we fix one person's seat, then the block can be in only one position (since the block is the entire table). Wait, but in this case, with n=3, fixing Pierre's seat, the block is the entire table. Then, the other two can be arranged in 2! ways, so total arrangements would be 2! = 2, which is correct because circular permutations of 3 is 2. So in that case, with n=3, the formula would give 2 * 2! * (something). Wait, no. Wait, in this case, the number of remaining people is 0, so it would be 2 * 2! * 0! = 2 * 2 * 1 = 4? Which is wrong. Wait, so maybe the formula is different.

Wait, in the case where n=3 and we fix Pierre's position, the block is the entire table, so there's only one position for the block (since it's the whole table). But earlier, I thought there were two positions, but that was in the case of n=8. Wait, maybe my previous logic was wrong.

Wait, let me re-examine. For n=8, fixing Pierre's seat. The block of three (Pierre, Rosa, Thomas) has to be consecutive. So, how many ways can that block be placed?

Since the table is circular, fixing Pierre at seat 1, the block can start at seat 1 and go clockwise to seats 1,2,3 or start at seat 1 and go counter-clockwise to seats 1,8,7. Therefore, two possible blocks. However, if we have a block of three seats, starting at seat 1, going clockwise, or starting at seat 1 going counter-clockwise. Therefore, two positions. However, if n=3, fixing Pierre at seat 1, the block would be seats 1,2,3 which is the entire table, but there's only one way to do that. So why in n=3, the number of positions is 1? Because if n=3, the block is the entire table, so there's only one possible block. Therefore, in general, for a block of size k in a circular table with n seats, fixing one person's position, the number of positions for the block is k? Or is it something else?

Wait, actually, if you fix one person's position, then the number of ways to place a block of size k that includes that person is equal to the number of ways to choose the rest of the block's positions. Since the block must include the fixed person, and be consecutive.

In a circular table with n seats, if we fix seat 1 as Pierre, then the block can be placed in two directions: clockwise or counter-clockwise. But if the block has size k, then depending on k, the number of positions may vary.

Wait, actually, given a fixed person, the number of ways to place a consecutive block of size k that includes that person is equal to 2 if the block is not the entire table. Wait, no. If the block is size k, then once you fix one person, the block can be placed in k positions? Wait, not sure.

Alternatively, think of the number of positions where a block of size k can be placed such that it includes the fixed person. For example, in a circular table with n seats, fixing seat 1. The number of blocks of size k that include seat 1 is equal to the number of ways to choose the other k-1 seats adjacent to seat 1 in one direction or the other. Wait, but if k is larger than 2, it's not just two.

Wait, let me take an example. If the block size is 3, and we fix seat 1, then the block can be seats 1,2,3 or seats 1,8,7 (assuming 8 seats). Therefore, two blocks. Similarly, if the block size is 4, fixing seat 1, the blocks would be 1,2,3,4 or 1,8,7,6. So two blocks. Wait, so for a block size of k, fixing one seat, the number of blocks that include that seat is 2? No, if k = n, then there's only 1 block. For k < n, it's two blocks? Wait, for example, n=8, k=3: 2 blocks. For n=8, k=2: the blocks that include seat 1 are seats 1,2 and seats 1,8. So two blocks. Similarly, for k=1, it's just seat 1. So in general, for a block size k, fixing seat 1, the number of blocks including seat 1 is 2 if k ≥ 2. Wait, but for k=2, yes, two blocks. For k=3, two blocks. Wait, but if the block size is k, and we fix one seat, the number of blocks is 2, unless k=1 or k=n. So in our problem, k=3, n=8. Therefore, fixing Pierre at seat 1, there are two blocks that include seat 1 and are of size 3. Therefore, two possible positions for the block. Then, for each of these blocks, the people inside can be arranged in different ways. However, since Pierre is fixed in seat 1, the rest of the block (Rosa and Thomas) can be arranged in 2! ways. So total arrangements for the block: 2 (positions) * 2! (internal arrangements) = 4.

Then, once the block is placed, the remaining 5 people can be arranged in 5! ways. Therefore, total number of arrangements is 4 * 5! = 4 * 120 = 480.

But earlier, the formula of treating the block as a single entity gave 5! * 3! = 720. Which is different. So which is correct?

Let me check with n=3 again. If we use this second approach: fixing Pierre's seat, there is 1 block (the whole table). The number of positions: 1. Then, internal arrangements: since the block is the entire table, and Pierre is fixed, the other two can be arranged in 2! ways. Then remaining people: zero, so 0! = 1. Therefore, total arrangements: 1 * 2! * 1 = 2, which is correct. So for n=3, the formula works. Then for n=8, the formula would be 2 * 2! * 5! = 2*2*120=480. But according to the first approach, treating the block as a single entity gave 5! * 3! = 720. Which one is correct?

Wait, let me think of another way. Suppose we don't fix Pierre's position. Let's try using the standard formula for circular permutations with a block.

In circular permutations, the formula when you have a block is (number of blocks - 1)! * internal permutations. But when the block is a single block, then number of blocks is 1. So (1 - 1)! * 3! = 1 * 6 = 6. Which is wrong, as discussed earlier. So that formula is wrong.

But when we fix one position, we get a different number. So why does fixing a position give a different answer?

Alternatively, maybe the correct formula is (number of entities - 1)! * internal permutations, but where entities are the blocks and the other individuals. However, if we don't fix positions, then we have to account for rotations. Wait, but in the case of fixing Pierre's position, we already accounted for rotational symmetry. So perhaps that approach is correct?

Alternatively, let me use another approach. The total number of circular permutations where the three people are together can be calculated by considering the following:

First, treat the three people as a single entity, so we have 6 entities (the block and the other 5 people). The number of circular permutations of these 6 entities is (6 - 1)! = 120. Then, multiply by the number of ways to arrange the three people within the block, which is 3! = 6. Therefore, total arrangements is 120 * 6 = 720. However, earlier when fixing Pierre's position, we got 480. Which is conflicting. What's going on?

Wait, perhaps when we fix Pierre's position, we are imposing an extra condition, so we get a different number. Wait, but both methods should give the same result. Therefore, there must be a mistake in one of the methods.

Wait, let's think of it in terms of linear arrangements. For linear arrangements, the number of ways where Pierre, Rosa, and Thomas are together is 6! * 3! (treating them as a block, so 6 entities, which can be arranged in 6! ways, multiplied by 3! for the block). But for circular arrangements, it's different.

But for circular arrangements, there's a formula that is similar to linear arrangements divided by n (because of rotations). Wait, but not sure.

Alternatively, consider that in circular arrangements, the number of arrangements where the three are together is equal to the number of linear arrangements where the three are together, divided by n. Because each circular arrangement corresponds to n linear arrangements (rotations). But let's check:

Number of linear arrangements where the three are together is (n - k + 1)! * k! So here, n=8, k=3. So (8 - 3 + 1)! * 3! = 6! * 3! = 720 * 6 = 4320. Wait, no, 6! is 720, times 3! is 720*6=4320. Then, if we divide by 8 (number of rotations), we get 4320 / 8 = 540. So that would be the number of circular arrangements.

But this contradicts both previous results of 480 and 720. Which one is correct?

Alternatively, maybe it's not divided by n, but something else. Wait, the formula for circular permutations where certain people are together.

Wait, another way: Let me think of it as arranging the block and the other people. In circular permutations, if we consider the block as a single entity, then we have 6 entities. The number of circular permutations is (6 - 1)! = 120. Then, multiply by 3! for the internal arrangements. So 120 * 6 = 720. However, this seems to be different from the linear approach divided by n.

But according to linear approach, we have 6! * 3! = 720 * 6 = 4320 linear arrangements. Then, if we consider circular arrangements, each circular arrangement corresponds to 8 linear arrangements (due to rotations), so number of circular arrangements would be 4320 / 8 = 540. But this contradicts with the previous result of 720.

So now, which is correct?

Alternatively, let's check with n=3 again. If n=3, the number of circular arrangements where all three are together is 2 (as before). Using the formula (n - k + 1)! * k! divided by n: (3 - 3 + 1)! * 3! / 3 = 1! * 6 / 3 = 6 / 3 = 2. Correct. Whereas the other approach, treating as block: (number of entities -1)! * internal permutations. Here, number of entities is 1, so (1 -1)! * 3! = 1 * 6 = 6. Which is wrong. Therefore, the formula of (number of entities -1)! * internal permutations is wrong? But in the linear case, the formula (n - k + 1)! * k! is correct. So perhaps for circular arrangements, the correct formula is [(number of entities -1)! * internal permutations]?

But for n=3, number of entities is 1, so [(1 - 1)! * 3!] = 6, which is wrong, but according to the division approach, we get correct result 2. Therefore, maybe the correct formula is [(number of entities -1)! * internal permutations] / something?

Wait, no. Wait, maybe the formula is different. Let me think again. In the case of circular permutations, when we have a block, we can think of it as follows:

If we have a block and other entities, the number of circular arrangements is equal to the number of linear arrangements of the entities divided by the number of rotations. Wait, but the number of linear arrangements is (number of entities)! * internal permutations. Then, to convert to circular arrangements, divide by the number of entities? Wait, no, because circular arrangements are (number of entities - 1)!.

Wait, perhaps the formula is (number of entities - 1)! * internal permutations, but this gives for n=3, 1 * 6 = 6, but actual answer is 2. Hence discrepancy.

Wait, so where is the mistake?

In the case of n=3, the formula (number of entities - 1)! * internal permutations is (1 -1)! * 3! = 6, but the correct number is 2. Therefore, the formula is wrong. Therefore, there must be another way.

Wait, but in linear arrangements, treating the block as a single entity, you get (number of entities)! * internal permutations. For n=3, that would be 1! * 3! = 6, which is the same as linear arrangements. But for circular arrangements, it's different. So how do they differ?

For circular arrangements, if we have a block, maybe the number is (number of entities - 1)! * internal permutations / something? Or is it different?

Wait, actually, in circular permutations, sometimes when you have symmetries, you have to adjust. For example, when considering necklaces with beads, sometimes you divide by symmetries.

But perhaps the standard formula is as follows:

When arranging objects around a circular table, if there are m objects, it's (m - 1)!.

If we have a block, and other objects, then the number of arrangements is (m - 1)! * (internal permutations of the block).

But in the case where the block is the only object, m=1, so (1 - 1)! * 3! = 1 * 6 = 6, but the actual number of circular arrangements is 2. Therefore, the formula is not correct in that case.

Wait, so what's different?

Ah! When we have a block, the internal permutations might be overcounted if the block has rotational symmetry. Wait, but in the case of n=3, the internal permutations are 3! = 6, but actual unique circular arrangements are 2. So perhaps the formula needs to divide by the number of rotations? But that seems complicated.

Alternatively, maybe when we treat the block as a single entity, we are overcounting because rotating the block around the table might lead to equivalent arrangements?

Wait, no, because the block is treated as an entity, and the rest of the table is arranged around it. Wait, this is getting too tangled.

Let me try to use the two methods I had earlier and see if I can find which is correct.

First method: fixing Pierre's position.

If Pierre is fixed, number of ways to arrange the rest is 2 * 2! * 5! = 4 * 120 = 480.

Second method: treating the block as a single entity, number of arrangements is (6 - 1)! * 3! = 120 * 6 = 720.

Third method: using linear arrangements divided by n: 6! * 3! / 8 = 720 * 6 / 8 = 540.

So three different answers. Which one is correct?

Let me think of a small example where I can compute manually. Let's take n=4, and want to find the number of circular arrangements where two specific people sit together.

Let's say people are A, B, C, D. Want A and B together. How many circular arrangements?

Total circular permutations for 4 people: 3! = 6.

Number of arrangements where A and B are together: ?

In circular table, fixing A's position. Then B can be to A's left or right: 2 possibilities. The other two people can be arranged in 2! ways. Total: 2 * 2 = 4. Which is equal to (number of entities - 1)! * internal permutations. Here, treating A and B as a block: number of entities is 3 (block + C + D). Then (3 - 1)! * 2! = 2! * 2 = 4. Which matches. So in this case, the formula works.

But if we use the alternative method of linear arrangements divided by n: linear arrangements where A and B are together: (n - k + 1)! * k! = (4 - 2 + 1)! * 2! = 3! * 2 = 6 * 2 = 12. Divided by 4: 12 / 4 = 3. Which is different from 4. So discrepancy here.

But manual calculation shows it's 4. Therefore, the correct formula is the first one: (number of entities - 1)! * internal permutations. So why the discrepancy with the third method?

Ah, because in the third method, when we calculate linear arrangements where the block is together, it's not just (n - k + 1)! * k! but actually (n - k + 1)! * k! ?

Wait, for n=4, k=2. Number of linear arrangements where A and B are together: treat them as a block, so 3 entities: block, C, D. These can be arranged in 3! ways, and the block can be in 2! ways, total 3! * 2! = 12. However, in circular arrangements, if we divide by n=4, we get 3, which is wrong. But manual calculation shows it should be 4. Therefore, the third method is flawed.

So why the difference? Because when converting from linear to circular, dividing by n is not always correct? Or maybe there's an adjustment needed?

Alternatively, in circular arrangements, the number of arrangements where the block is together is equal to (number of entities - 1)! * internal permutations. Which in this case works. For n=4, entities = 3, so (3-1)! * 2! = 2 * 2 = 4. Which is correct.

Therefore, maybe the formula is correct, and the third method was wrong. So why did that happen?

Because when I divided by n, I assume that each circular arrangement corresponds to n linear arrangements, but perhaps when the block is considered, some symmetries are counted differently.

Therefore, trusting the formula (number of entities -1)! * internal permutations, which worked for n=4 and n=3 (when I adjust the numbers), let me go back to the original problem.

Original problem: 8 people, with a block of 3 (Pierre, Rosa, Thomas). So number of entities: 8 - 3 + 1 = 6. Therefore, number of circular arrangements is (6 - 1)! * 3! = 5! * 6 = 120 * 6 = 720.

But earlier, when fixing Pierre's position, I got 480. Why is there a discrepancy?

Wait, if I use the formula for fixing Pierre's position: fixed Pierre, then the number of ways is 2 * 2! * 5! = 480. But according to the entity method, it's 720. Which is different.

Wait, but if I don't fix Pierre's position, the total number should be 720. But when I fix Pierre's position, I should get 720 / 8 = 90? Wait, no. Wait, fixing one position in circular permutations reduces the count by a factor of n. Wait, no. If total number is 720, then fixing one person's position would result in 720 / 8 = 90. But according to the other method, fixing Pierre's position, I have 480. Which is way bigger than 90. So something's wrong.

Wait, clearly there is a miscalculation in the fixing position method. Let me re-examine that.

If we fix Pierre's position, then we are effectively converting the circular problem into a linear one, with Pierre's position fixed. So the number of arrangements should be equal to the total number divided by n (the number of seats), but in this case, it's not matching.

Let me redo the fixing method. Let's fix Pierre at seat 1. Now, we need to arrange the remaining 7 people around the table with the condition that Rosa and Thomas are next to Pierre or to each other in a block.

Wait, no, the condition is that all three (Pierre, Rosa, Thomas) must sit together. Since Pierre is fixed at seat 1, the block of three must include seat 1. Therefore, the block can be either seats 1,2,3 or seats 1,8,7. Therefore, two possible positions for the block.

For each of these positions, the block contains Pierre, plus two others. Since Pierre is fixed, the other two can be arranged in 2! ways. So for each block position, 2! arrangements. So total for blocks: 2 * 2! = 4.

Then, the remaining 5 people can be arranged in the remaining seats. How many seats are left? Total seats are 8. The block takes 3 seats, so remaining seats: 8 - 3 = 5. So those 5 people can be arranged in 5! ways. Therefore, total arrangements with Pierre fixed: 4 * 5! = 4 * 120 = 480.

But if the total number of arrangements without fixing is 720, then fixing Pierre's position should give 720 / 8 = 90. But here, we get 480. Which is not matching. So there's a contradiction.

Wait, this suggests that my first method using entities is wrong, and the fixing method is correct? Or vice versa?

Wait, perhaps my fixing method is wrong. Let me think again. If I fix Pierre's position, the total number of arrangements where the block is together should be equal to total number of circular arrangements with the block together divided by 8 (since fixing one position accounts for rotational symmetry). So if total number is X, then fixing Pierre's position would give X / 8. But according to fixing method, I get 480, so X would be 480 * 8 = 3840, which is way higher than the entity method result of 720.

But earlier with n=4, fixing a position worked. Let me check with n=4, k=2. Total number of circular arrangements where two people are together: 4. If I fix one person's position, say A, then number of arrangements is 2 * 1! * 2! = 2 * 1 * 2 = 4. Wait, no, let me redo it.

For n=4, people A, B, C, D. Fix A at seat 1. Block of A and B must be together. The block can be seats 1,2 or seats 1,4. For each, B can be at 2 or 4. Since the block includes A and B, and A is fixed at seat 1, then B can be at 2 or 4. So two possibilities. Then the remaining two people (C and D) can be arranged in seats 3 and the remaining seat. Wait, if block is seats 1,2, then remaining seats are 3 and 4. But seat 4 is adjacent to seat 1. Wait, if block is seats 1,2, then seats 3 and 4 are remaining. So C and D can be arranged in 2! ways. Similarly, if block is seats 1,4, then remaining seats are 2 and 3, which can be arranged in 2! ways. So total arrangements: 2 (block positions) * 2! (arrangements of remaining) = 2 * 2 = 4, which matches the total circular arrangements (which is 4). So in this case, fixing position gives the correct result, and the total number is 4. If I had used the entity method: number of entities = 4 - 2 + 1 = 3. (3 - 1)! * 2! = 2! * 2 = 4. Correct.

So why in the original problem, fixing Pierre's position gives 480, while the entity method gives 720? Let's check if there's an error in the fixing method. For n=8, fixing Pierre's position, number of arrangements where the block is together is 4 * 5! = 480. If we trust this, then total number of circular arrangements should be 480 * 8 = 3840. But according to the entity method, it's 720. Which is different.

But for n=4, the entity method and fixing position method gave the same result. So why in n=8, they differ? There must be a miscalculation in one of the methods.

Wait, in the n=4 example, the number of entities was 3, leading to (3-1)! * internal permutations = 2! * 2 = 4, which matches the fixing method. So in that case, the total number was 4, and with fixing position, also 4. So why for n=8, fixing gives 480, while the entity method gives 720?

Wait, let me recast the entity method for n=8. Number of entities is 6 (block + 5 people). Circular arrangements: (6-1)! * 3! = 120 * 6 = 720. So total arrangements is 720. If I fix Pierre's position, the number should be 720 / 8 = 90. But according to fixing method, it's 480. Therefore, there's a contradiction.

Therefore, one of the methods must be wrong. Let me try to find the error.

Wait, when I fixed Pierre's position, I considered that the block could be in two positions (seats 1-2-3 or 1-8-7). But maybe there are more positions?

Wait, no. With n=8 and block size 3, fixing Pierre at seat 1, the block must include seat 1 and be consecutive. The only possible blocks are seats 1-2-3 and seats 1-8-7. So two blocks. So why is there a discrepancy?

Wait, maybe the error is that when I fixed Pierre's position, I didn't account for the fact that the block can be rotated? Or is there a miscalculation in the internal arrangements?

Wait, if the block is in seats 1-2-3, then seat 1 is Pierre, and seats 2 and 3 are Rosa and Thomas. They can be arranged in 2! ways. Similarly, if the block is in seats 1-8-7, seat 1 is Pierre, seats 8 and 7 are Rosa and Thomas, also 2! ways. So total for the block: 2 * 2! = 4. Then remaining 5 people: seats 4,5,6,7,8? Wait, no. If the block is seats 1-2-3, then remaining seats are 4,5,6,7,8. But seat 8 is adjacent to seat 1. Wait, no, if the block is seats 1-2-3, then the remaining seats are 4,5,6,7,8. Wait, but seat 8 is separate. Yes, so those 5 seats can be filled with the remaining 5 people in 5! ways. Similarly, if the block is seats 1-8-7, remaining seats are 2,3,4,5,6. Which are also 5 seats. So 5! ways. So total is 2 * 2! * 5! = 4 * 120 = 480.

But according to entity method, it should be 720. So why the difference? Let me think if there is another way to place the block.

Wait, if we don't fix Pierre's position, but consider all possible positions for the block, how many are there? For a block of size 3 in a circular table of 8 seats, the number of blocks is 8 (each starting at a different seat). But since rotations are considered the same, if we don't fix any position, the number is 8 / 8 = 1? No, that's not right.

Actually, the number of distinct blocks in circular table is n (number of seats), but since rotations are considered the same, but in our case we are counting all possible blocks (which would be n, but considering rotations as distinct? No, in the entity method, we treat the blocks as distinct entities.

Wait, this is getting too confusing. Maybe I should trust the fixing method, since it worked for n=4 and it seems logical.

Alternatively, think of the entity method. If we have 6 entities (block + 5 people), then number of circular arrangements is (6-1)! * 3! = 120 * 6 = 720. But if we fix one position, say, the block's position, then we have to adjust.

Wait, let me think: if we fix the block's position, then the number of arrangements is 3! * 5! (since block is fixed, internal arrangements are 3! and remaining people are 5!). But since we fixed the block's position, we don't have to divide by anything. But how many ways to fix the block's position?

Wait, in circular table, fixing the block's position would mean that we can rotate the table so that the block is in a specific spot. So if we fix the block at a specific position, then the number of arrangements is 3! * 5! = 6 * 120 = 720. But since there are 8 seats, and the block takes up 3, the number of positions for the block is 8, but since rotations are considered the same, fixing the block's position accounts for all rotations. Wait, no, fixing the block's position means we're considering rotational symmetry already. Therefore, the total number of circular arrangements is 3! * 5! = 720. This matches the entity method.

But earlier, when I fixed Pierre's position, I got 480. Why? Because fixing Pierre's position is a different constraint. If instead of fixing the block's position, I fix Pierre's position, it affects the count. Let me see.

If I fix the block's position, then Pierre can be in any position within the block. But if I fix Pierre's position, it restricts the block's position. Hence, the different counts.

Therefore, the correct answer is 720, from the entity method. But why does fixing Pierre's position give 480? Probably because by fixing Pierre's position, I'm not accounting for all possible rotations where the block is placed relative to Pierre.

Wait, but in fixing Pierre's position, I considered all possible placements of the block that include Pierre. So why is it different?

Let me think: total number of circular arrangements is 720. If I fix Pierre's position, how many arrangements are there? It should be 720 / 8 = 90. But according to my calculation, it's 480. So there is a miscalculation.

Wait, let me recast the fixing Pierre's position method. Suppose we fix Pierre at a seat. The total number of arrangements where the three are together is equal to the number of ways to arrange the block and the others.

But if I fix Pierre's position, then the block is determined by where Rosa and Thomas are. But how many ways are there?

Alternatively, think of it as: once Pierre is fixed, the block of three can be arranged in 2 * 2! ways (as before), and the rest arranged in 5! ways, giving 4 * 120 = 480. But total number of arrangements with Pierre fixed is 480. But total number of circular arrangements is 720. So if I fix Pierre's position, the number should be 720 / 8 = 90, but it's 480. So clearly, something is wrong.

Wait, but if I fix Pierre's position, the total number of arrangements without any restrictions is (8-1)! = 5040? No, wait. If we fix Pierre's position, then the number of circular arrangements is (8-1)! / 8 = 7! / 8 = 5040 / 8 = 630? No, wait: when you fix one person's position in a circular arrangement, the number of distinct arrangements is (n-1)! So for 8 people, fixing one person, it's 7! = 5040? No, no! Wait, if you fix one person's position, the number of arrangements is (n-1)! / n * n = (n-1)! Wait, no. Let me clarify.

When you fix one person's position in a circular arrangement, you're effectively converting it into a linear arrangement. The number of distinct circular arrangements is (n-1)! So if you fix one person's position, the number of linear arrangements is (n-1)! So for n=8, fixing Pierre's position, the number of arrangements is 7! = 5040? No, that can't be. Wait, no. Wait, for n people, the number of circular arrangements is (n-1)! If you fix one person's position, you're removing the rotational symmetry, so the number of arrangements is (n-1)! / n * n = (n-1)!? Wait, I'm confused.

Actually, fixing one person's position in a circular table makes it equivalent to arranging the remaining (n-1) people linearly. So the number of arrangements is (n-1)!.

Wait, for n=8, fixing Pierre's seat, the number of arrangements is 7! = 5040. But earlier, when I calculated the number of arrangements where the block is together, I found 480. So 480 / 5040 = 480 / 5040 = 4/42 = 2/21. Which is the probability that the three are together, which seems plausible?

But if the total number of arrangements is 720, and the number of arrangements with block together is 720, then fixing Pierre's position would give 720 / 8 = 90. But according to fixing method, it's 480. So clearly, there is a miscalculation.

Wait, let me check what is the total number of arrangements with the block together. If it's 720, and total number of circular arrangements is 7! = 5040, then the probability is 720 / 5040 = 1/7. But if I fix Pierre's position, the probability that the block is together is 480 / 7! ? 480 / 5040 = 4/42 = 2/21 ≈ 0.095, while 1/7 ≈ 0.142. So they don't match. So one must be wrong.

Alternatively, let me calculate the total number of circular arrangements where the three are together correctly.

Using the entity method: 5! * 3! = 720. Total circular arrangements: 7! = 5040. So the probability is 720 / 5040 = 1/7 ≈ 0.142.

Now, using the fixing method: fixing Pierre's position, number of such arrangements is 480, and total arrangements with Pierre fixed is 7! = 5040? No, wait: if we fix Pierre's position, the total number of arrangements is (8-1)! = 5040? No, no! Wait, if you fix one person's position in a circular table, the number of distinct arrangements is (n-1)! So for n=8, fixing Pierre's seat, it's 7! = 5040? Wait, no! Wait, no. Wait, when you fix one person's position, you are effectively making the problem linear. The number of arrangements is (n-1)! So for n=8, it's 7! = 5040? No! Wait, no. Wait, the number of circular permutations is (n-1)! So if you fix one person, the number of arrangements is (n-1)!

Wait, for example, for n=3, fixing one person, the number of arrangements is 2! = 2, which is correct. For n=4, fixing one person, it's 3! = 6, which is correct. So yes, for n=8, fixing one person, the number of arrangements is 7! = 5040. But earlier, I thought it was (n-1)! = 5040, which is correct.

But according to fixing method, the number of arrangements with the block together is 480, so the probability is 480 / 5040 = 4/42 = 2/21 ≈ 0.095, while the correct probability is 1/7 ≈ 0.142. Hence, discrepancy.

Therefore, there must be an error in the fixing method. Let me find it.

When I fixed Pierre's position, I assumed that the block could be in two positions (seats 1-2-3 or 1-8-7), each with 2! internal arrangements, and the rest arranged in 5! ways. But perhaps this is missing some arrangements?

Wait, maybe there are more ways to position the block? Or maybe the internal arrangements are not just 2!?

Wait, if the block is fixed to include Pierre, but the block can be in more positions? No, for n=8, block size 3, fixing Pierre at seat 1, there are only two possible blocks: 1-2-3 and 1-8-7. So that seems correct.

Wait, but perhaps the internal arrangements are not just 2!?

Wait, in each block, Pierre is fixed at seat 1. The other two seats in the block are occupied by Rosa and Thomas. So for the block 1-2-3, seats 2 and 3 are Rosa and Thomas, which can be arranged in 2! ways. Similarly, for block 1-8-7, seats 8 and 7 are Rosa and Thomas, which can be arranged in 2! ways. So total of 2 * 2! = 4. Then the rest of the seats, 5 people, arranged in 5! ways. So total 4 * 120 = 480.

But if the correct total number is 720, then where is the missing 720 - 480 = 240?

Ah! Wait, perhaps the block can be in more positions? For example, if the block is not required to include Pierre's seat? Wait, no! The condition is that Pierre, Rosa, and Thomas all sit together. Since Pierre is fixed, the block must include Pierre. So the block must include seat 1.

Wait, but what if the block is somewhere else, but still includes Pierre? No, because Pierre is fixed at seat 1. So the block must include seat 1. Therefore, only two possible blocks.

Wait, but what if the block is placed such that Pierre is in the middle of the block? For example, if the block is seats 8, 1, 2. Which is the same as seats 1-2-8, but that is already counted as block 1-8-7 or 1-2-3? No. Wait, block 1-8-7 is seats 1, 8, 7. Block 1-2-3 is seats 1, 2, 3. So if the block is seats 8, 1, 2, that's the same as block 1-2-8, which is the same as block 1-2-3? No, because in a circular table, the block seats 1, 2, 3 is consecutive, and block 1, 8, 7 is also consecutive. But block 1, 2, 8 is not consecutive, since 2 and 8 are not adjacent. So no, that's not a valid block.

Therefore, only two possible blocks that include seat 1. So why is there a discrepancy?

Wait, let me think differently. Let me calculate the number of ways where Pierre, Rosa, and Thomas are together, with Pierre fixed at seat 1.

The block of three can start at seat 1 and extend clockwise or counter-clockwise. For each direction, the block is three seats. So, for each direction, the block is determined, and within the block, Pierre is fixed, and the other two can be arranged. But perhaps there are more ways to arrange the block?

Wait, if the block is allowed to be in any position, but includes Pierre, then there are more positions? No, because Pierre is fixed, and the block must be consecutive. So only two positions.

Wait, here's a thought: perhaps the block can include Pierre in different positions. For example, if the block is three seats, and Pierre is in the middle or at the end. But since we're fixing Pierre at seat 1, he can only be at the start of the block, or in the middle? Wait, no, if the block is three seats, and Pierre is at seat 1, then he is at the start of the block if the block is 1-2-3, or at the end of the block if the block is 7-8-1. Wait, but in this case, the block is 7-8-1, which includes seat 1. So in this case, Pierre is at the end of the block. But in this case, the block is 7-8-1, which is consecutive. So, but in this case, the block can be arranged such that Pierre is at the end, and Rosa and Thomas are at 7 and 8. But this is already counted in the block positions 1-8-7. So it is already included in the two block positions.

Therefore, no additional block positions.

So why is the count different?

Wait, let me think about the number of ways to have Pierre, Rosa, Thomas together with Pierre fixed. Let's think of it as: the number of ways where Pierre is at seat 1, and Rosa and Thomas are adjacent to him or to each other.

Wait, but they must all be together. So the three of them must occupy three consecutive seats, with Pierre at seat 1. So the three seats must be 1-2-3 or 1-8-7. In each of these cases, the other two people (Rosa and Thomas) can be arranged in 2! ways. Then, the remaining 5 people can be arranged in 5! ways. So total 2 * 2! * 5! = 480.

But if the correct answer is 720, then where is the error?

Wait, perhaps fixing Pierre's position is not the right approach here, and the entity method is correct? Or maybe the entity method overcounts?

Wait, let me think of it this way: in the entity method, we treat the block as a single entity, giving us 6 entities. Number of circular arrangements is (6-1)! * 3! = 120 * 6 = 720. But if we calculate it by fixing one person, we get 480.

But if I don't fix any positions, and calculate using the entity method, it should be correct. Therefore, there must be a mistake in the fixing method.

Let me try a different approach: let's calculate the number of circular arrangements where Pierre, Rosa, and Thomas are together by considering all possible positions of the block.

In a circular table with 8 seats, the number of ways to choose a block of 3 consecutive seats is 8 (since it's circular). For each block, we can arrange Pierre, Rosa, Thomas in 3! ways. Then, arrange the remaining 5 people in the remaining 5 seats in 5! ways.

However, in circular arrangements, this counts each arrangement multiple times. Specifically, each arrangement is counted once for each rotational position of the block. Wait, but since the block is fixed in position, and we are counting all blocks, this may not account for rotational symmetry.

Wait, no. If we calculate it as 8 (number of blocks) * 3! * 5! = 8 * 6 * 120 = 5760. But this is the number of linear arrangements, because each block is considered distinct, and rotations are not accounted for. To convert to circular arrangements, we need to divide by the number of rotations, which is 8. So 5760 / 8 = 720. Which matches the entity method!

Ah! So this is the correct way. So the error in the fixing method was that I fixed Pierre's position and didn't consider all possible blocks. By considering all blocks and dividing by the number of rotations, we get 720.

Therefore, the correct answer is 720. So why did the fixing method give 480? Because when I fixed Pierre's position, I restricted the block to only those that include Pierre, but in reality, the block can be anywhere, and fixing Pierre's position still allows for other blocks that include Pierre. Wait, no. Wait, in this calculation, when we fix Pierre's position, we are not considering all possible blocks, but only those that include Pierre. But if we want to calculate the total number of arrangements where the block includes Pierre, we should calculate it as follows: number of blocks including Pierre is 8 (total blocks) * ... wait, no.

Actually, the number of blocks that include Pierre is equal to the number of blocks where Pierre is in the block. Since each block of 3 consecutive seats contains 3 people, and there are 8 blocks, the total number of blocks is 8. How many of these blocks include Pierre? For each block, it can start at any seat. For example, block starting at seat 1: includes seats 1-2-3. Block starting at seat 2: seats 2-3-4. ... Block starting at seat 8: seats 8-1-2. So, how many of these blocks include seat 1? Blocks starting at seat 8 (includes 8-1-2), seat 1 (1-2-3), and seat 2 (2-3-4) does not include 1. Wait, no. Block starting at seat 8 includes seat 8, 1, 2. Block starting at seat 1 includes 1, 2, 3. Block starting at seat 8 and 1 include seat 1. So total of 2 blocks that include seat 1. Therefore, out of 8 total blocks, 2 include seat 1. So if we fix Pierre's position at seat 1, the number of blocks that include him is 2. So, for each of these 2 blocks, we can arrange the three people in 3! ways, and the rest in 5! ways. So total is 2 * 3! * 5! = 2 * 6 * 120 = 1440. But this is the number of arrangements where the block includes Pierre, but since we fixed Pierre's position, these are the number of arrangements where the block includes Pierre. But since in total, there are 8 blocks, each corresponding to a different position, but fixed Pierre's position restricts it.

Wait, but this seems to give 1440, which is higher than the total number of circular arrangements (5040). But no, because when we fix Pierre's position, the total number of arrangements is 7! = 5040, and the number of arrangements where the block includes Pierre is 1440, which is 1440 / 5040 = 1440 / 5040 = 12/42 = 2/7, which is different from the earlier probability of 1/7. Wait, but this is contradictory.

This shows that there is confusion in the fixing method. It seems that the correct answer is 720, as per the entity method and the block counting divided by rotations. Therefore, I must have made a mistake in the fixing method.

Given that, and since the entity method and the block counting method both give 720, I will trust that the correct answer is 720.

So, the number of ways is 5! * 3! = 120 * 6 = 720.

Therefore, the final answer is \boxed{720}.

**Final Answer**
\boxed{720}

To determine how many ways 8 people can sit around a round table **such that** Pierre, Rosa, and Thomas all sit **together**, we proceed by considering circular permutations with the specific constraint that a particular group (of size 3) must **be seated consecutively**.

---

### Step 1: Understand Circular Permutations with Group Constraints

In standard **circular permutations**, the number of distinct arrangements of $ n $ people is $ (n-1)! $. This is because **rotations** of the table are considered **identical**, so we fix one person’s position to eliminate overcounting.

However, when a **group** of people must all sit **together**, we can treat this group as a **single "block"**. This effectively reduces the problem to arranging this block with the remaining individuals.

---

### Step 2: Break the Problem into Two Parts

- **Group as a block**:
  We consider the 3 people (Pierre, Rosa, Thomas) as a block, which can be arranged internally in $ 3! $ ways.

- **Remaining people**:
  The remaining 5 people plus the "block" form $ 6 $ entities to be arranged in a circle.

---

### Step 3: Count the Circular Arrangements of the 6 Entities

Since we're dealing with **circular permutations** of these 6 entities, the number of arrangements is:

$$
(6 - 1)! = 5! = 120
$$

---

### Step 4: Multiply by Internal Arrangements of the Block

The block of 3 people can be arranged among themselves in $ 3! = 6 $ ways.

---

### Step 5: Final Calculation

$$
\text{Total number of valid arrangements} = 5! \times 3! = 120 \times 6 = \boxed{720}
$$

---

### Final Answer

$$
\boxed{720}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways 8 people can sit around a round table, but with the condition that Pierre, Rosa, and Thomas all want to sit together. Hmm, circular permutations can be a bit tricky because rotations are considered the same. Let me think through this step by step.

First, I remember that for circular permutations of n people, the number of distinct arrangements is (n-1)! instead of n! because rotations are considered the same. But here, there's a condition: three specific people must sit together. So I need to adjust for that.

Let me recall, when dealing with groups that must stay together in circular permutations, we can treat the group as a single entity or "block." So if Pierre, Rosa, and Thomas must sit together, maybe I can consider them as one block and then arrange the blocks around the table. But since it's a circular table, I need to be careful with how I count the arrangements.

So, normally, if we have 8 people, and we're grouping 3 of them into a block, we're effectively creating (8 - 3 + 1) = 6 "items" to arrange around the table: the block and the other 5 individuals. But wait, actually, the total number of items would be 1 block + 5 other people = 6 items. However, in circular permutations, the number of ways to arrange n items is (n-1)! So here, arranging 6 items around a circular table would be (6-1)! = 5!.

But then, within the block, the three people can be arranged among themselves in 3! ways. So total arrangements would be 5! * 3!.

But let me check if that's correct. Let me think again. If it's a circular table, fixing the block's position to account for rotations? Wait, but when we treat the block as a single entity, do we still need to adjust for rotations? Hmm.

Alternatively, sometimes people use the formula for circular permutations with a group: fix the position of the block to eliminate rotational symmetry. Let me think. For example, if you have a block of people, you can fix the block in a certain position, then arrange the rest relative to it. But since rotations are considered the same, fixing the block's position would account for that.

But if I consider the block as a single entity, then arranging the block and the other 5 people around the table is (number of entities - 1)! So (6 - 1)! = 5! as before. Then multiply by the number of internal arrangements of the block, which is 3!.

So that would give 5! * 3! = 120 * 6 = 720. Wait, but let me check if that's correct.

But wait, let me think again. Suppose there are no restrictions. The total number of circular permutations is 7! / 8 = 5040 / 8 = 630? Wait, no. Wait, circular permutations formula is (n-1)! So for 8 people, it's 7! = 5040. Wait, no! Wait, no. Wait, hold on. Wait, (n-1)! is the formula. So for 8 people, it's 7! = 5040. But if we have a restriction that three people must sit together, then the method I mentioned before would be treating them as a block, so 6 entities, so (6-1)! = 5! arrangements, multiplied by 3! for the block. So 5! * 3! = 120 * 6 = 720. But 720 is less than 5040, which seems plausible.

Wait, but is there another way to think about it?

Alternatively, suppose we fix one person's position to remove rotational symmetry. Let me try that approach. Let me fix Pierre's position. Since rotations are considered the same, we can fix Pierre at a specific seat. Then, we need to arrange the remaining 7 people around the table. But since Pierre, Rosa, and Thomas must all sit together, Rosa and Thomas have to be adjacent to Pierre or to each other?

Wait, if Pierre is fixed, then the block of Pierre, Rosa, and Thomas can be arranged in different ways. Let me think. If we fix Pierre's position, then the block can be considered as Pierre sitting in a specific seat, and Rosa and Thomas sitting next to him. However, since the block can be arranged in different orders.

But since we fixed Pierre's position, the block can be arranged in two directions: for example, Pierre, Rosa, Thomas or Pierre, Thomas, Rosa. So, the block can be arranged in 2! ways? Wait, no, since within the block, there are 3 people, so once Pierre is fixed, Rosa and Thomas can be arranged in 2! ways? Wait, but if the entire block is considered, maybe we need to consider the block's position.

Wait, if Pierre is fixed, then the block can be placed in different positions. Wait, but since the table is circular and we fixed Pierre's position, the block can be arranged either clockwise or counter-clockwise around Pierre?

Wait, this seems confusing. Let me think again. If we fix Pierre's position, then the remaining people are Rosa, Thomas, and 5 others. The condition is that Rosa and Thomas must be adjacent to Pierre or to each other? Wait, no, the condition is that all three must sit together. So the block of Pierre, Rosa, and Thomas must be together. Since Pierre is fixed, the block can be arranged in two directions: either going clockwise or counter-clockwise from Pierre.

Wait, but actually, if Pierre is fixed, then the block can be thought of as a sequence of three people: Pierre, then Rosa and Thomas next to him. But since there's three people in the block, if Pierre is fixed, the block can be arranged in two different ways? Or perhaps more?

Alternatively, fixing Pierre's position, the block of three can be arranged in 2! ways? Wait, no. Let me think. If we fix Pierre in a seat, then we can arrange Rosa and Thomas in the seats adjacent to Pierre. But since it's a block of three people, they need to occupy three consecutive seats. If Pierre is fixed, then the block can be placed either to his left or to his right? Wait, but in a circular table, if Pierre is fixed, then the seats are ordered clockwise. So if we fix Pierre at seat 1, then the block could occupy seats 1, 2, 3 or seats 1, 8, 7? Wait, but if we fix Pierre at seat 1, then the block must include seat 1. So the block can be seats 1, 2, 3 or seats 1, 8, 7. But if we consider that the block must be three consecutive seats, then fixing Pierre at seat 1, the block can be either seats 1, 2, 3 or 1, 8, 7. However, if we consider the entire block as three consecutive seats, then once Pierre is fixed, the block can be placed in two different linear arrangements? Wait, but actually, the three people can be arranged in different permutations.

Wait, perhaps this approach is overcomplicating. Let me check the standard approach for circular permutations with a block.

In linear permutations, if we have a block, we treat it as a single entity, so the number of arrangements is (n - k + 1)! * k! where n is total people and k is the size of the block. But for circular permutations, it's different because of rotational symmetry.

But here's a method: For circular permutations with a block, you can fix the position of the block to remove rotational symmetry. So if you have a block of size k, you can fix the block in a specific position, then arrange the remaining people. However, fixing the block's position would mean that the number of arrangements is (total number of items - 1)! * internal arrangements.

Wait, maybe I should think of the block as an entity. So total entities: 1 block + 5 other people = 6 entities. In circular permutations, arranging n entities around a circular table is (n-1)! So here, (6-1)! = 5! Then, multiply by the number of ways to arrange the people within the block, which is 3!.

Therefore, total number of arrangements is 5! * 3! = 120 * 6 = 720. So that would be the answer. But let me check if this is correct.

Alternatively, another approach: total number of circular arrangements where the three people are together is equal to the number of linear arrangements where the three are together, divided by something? Wait, no, because circular vs linear is different.

Alternatively, suppose we consider the three people as a single entity. Then, in circular permutations, the number of ways to arrange the entities is (number of entities - 1)! So that's (6 - 1)! = 5! Then multiply by the permutations within the block: 3!. Therefore, 5! * 3! = 720. So that seems to be the same as before.

But let me check with an example. Let's say there are fewer people. Suppose there are 3 people, and they all must sit together. Then, how many arrangements? Well, since it's a circular table, there's only 1 arrangement. Using the formula: number of entities is 1 (the block), so (1 - 1)! = 0! = 1, multiplied by 3! = 6. Wait, that gives 6, which is wrong. Wait, that can't be. So clearly, this formula isn't correct. So my previous reasoning is flawed.

Wait, what's the problem here? If there are 3 people sitting around a round table, and they must all sit together, there's actually only 1 way, because rotations are considered the same. However, according to the formula (number of entities - 1)! * internal permutations, which would be (1 - 1)! * 3! = 1 * 6 = 6. Which is wrong. Therefore, my formula is incorrect.

So there's a mistake in my reasoning. Let me figure out why.

Ah, so when the entire group is the block, then treating it as a single entity, but in circular permutations, when you have one entity, the number of arrangements is (1 - 1)! = 1, which is correct, but then multiplied by 3! gives 6, which is wrong. So clearly, there's an error here.

Wait, maybe when we have only one block, the number of circular arrangements is 1, as there's only one way to arrange a single block around a table. But if you multiply by the internal permutations, then 1 * 3! = 6. But in reality, the number of circular arrangements where all 3 people are sitting together is 1 (since rotations are considered the same), but if you consider rotations as different, then it would be 2 (since in a circular table, there are (n-1)! arrangements, so for n=3, it's 2. Wait, hold on. Let me clarify:

Wait, if we have 3 people around a circular table, the number of distinct seating arrangements is 2, not 1. Wait, no. Wait, for n people, circular permutations are (n-1)! So for 3 people, it's 2. However, if we consider that all three must sit together, but they are already seated together. So the number of arrangements is 2. However, according to the formula, treating the three as a single block, the number of arrangements is (number of blocks -1)! * internal permutations. Which would be (1 - 1)! * 3! = 1 * 6 = 6. Which is wrong.

So that approach is wrong. Therefore, my initial reasoning was incorrect. So why is that?

Ah, maybe because when we consider the block, if we have only one block, then the number of arrangements is 1 (for the block) but the internal permutations are 3! but in reality, fixing the block's position, the internal permutations are 3! but since rotations are already accounted for by fixing the block's position. Wait, but how?

Alternatively, maybe the formula is different. Maybe when we have a single block, the number of circular arrangements is (number of entities - 1)! * internal permutations. But if there is only one entity, then (1 - 1)! * 3! = 1 * 6 = 6. However, in reality, the number is 2. So clearly something is wrong here.

Therefore, there's a mistake in my approach. Let me think again. What's the correct way to count circular permutations with a block?

Let me refer back to standard methods. In circular permutations, when we have a group that must stay together, we can use the following approach:

First, consider the group as a single "super person". Then, the number of ways to arrange the "super person" and the other individuals around the table is (number of entities - 1)!.

However, this is only if we consider that the "super person" can be rotated. But actually, when we treat the group as a single entity, the number of circular arrangements is (number of entities - 1)!.

However, then we have to multiply by the number of ways to arrange the people within the group.

But in the case where the group is the entire set, this would lead to (1 - 1)! * 3! = 1 * 6 = 6. But for n=3, the correct number is 2. So why is this discrepancy?

Ah! Because in the case where the entire group is the block, we have to consider that in circular permutations, when we fix the block's position, the internal arrangements are counted as permutations. However, if we don't fix the block's position, then the formula is different. Wait, actually, I think the formula is correct when there are multiple blocks, but perhaps not when there's only one block?

Wait, no. Wait, let me think again. Suppose we have n people, and we want to arrange them around a circular table with a specific block. Let's say we have a block of size k. So total entities: m = n - k + 1. Then, arranging m entities around a circular table is (m - 1)! and multiplying by k! for internal arrangements. However, in the case where m = 1 (i.e., n = k), this gives (1 - 1)! * k! = 1 * k! = k! which is wrong because for circular permutations it's (n - 1)!.

Wait, so there's a contradiction here. So clearly, the formula is not applicable when the entire set is a block. Therefore, my previous reasoning is wrong. Therefore, I need another approach.

So what is the correct way?

Let me think. Let me recall that when dealing with circular permutations with groups, the formula is similar to linear permutations but adjusted for circularity. For linear permutations, the number would be (n - k + 1)! * k! So for example, for 3 people in a line, it's (3 - 3 + 1)! * 3! = 1! * 6 = 6, which is correct. But for circular permutations, perhaps it's different.

Wait, another approach: For circular permutations, we can fix one person's position to remove rotational symmetry. Let me try that.

Let me fix Pierre's position. Since rotations are considered the same, fixing Pierre at a specific seat. Then, the problem reduces to arranging the remaining 7 people around the table with the condition that Rosa and Thomas are adjacent to Pierre or to each other? Wait, no, the condition is that all three (Pierre, Rosa, Thomas) must sit together. Since Pierre is fixed, the block of three must include Pierre and the two others. So, the block is Pierre, Rosa, and Thomas, and they need to be consecutive.

Since Pierre is fixed, the block can be arranged in two directions: either clockwise or counter-clockwise from Pierre. Wait, but how does that work?

Let me visualize the table. Let's fix Pierre at seat 1. Then, the block of three can occupy seats 1, 2, 3 or seats 1, 8, 7. These are the only two possible blocks that include seat 1 and are consecutive. Therefore, there are two possible positions for the block.

But within each block, the three people can be arranged in different orders. However, since Pierre is fixed at seat 1, the block's arrangement is determined by the positions of Rosa and Thomas.

Wait, for example, if the block occupies seats 1, 2, 3, then seat 1 is Pierre, so seats 2 and 3 must be Rosa and Thomas in some order. Similarly, if the block occupies seats 1, 8, 7, then seat 1 is Pierre, seat 8 and seat 7 must be Rosa and Thomas in some order. Therefore, for each block position (two possibilities), we have 2! arrangements for Rosa and Thomas. Therefore, total arrangements for the block is 2 (positions) * 2! (internal arrangements). Wait, but Pierre is fixed, so is that correct?

Wait, if the block can be in two positions, and in each position, the other two people can be arranged in 2! ways, so total of 2 * 2! = 4 ways. Then, once the block is placed, the remaining 5 people can be arranged in the remaining 5 seats in 5! ways. Therefore, total number of arrangements would be 4 * 5!.

But let me check that. Wait, if we fix Pierre's seat, then the block can be in two positions (left or right side). For each position, Rosa and Thomas can switch places, so 2! ways. Then, the remaining 5 people can be arranged in the remaining 5 seats. Therefore, total is 2 * 2! * 5!.

But then 2 * 2! * 5! = 2 * 2 * 120 = 480. But let me check with the previous formula (which gave 720). Which one is correct?

Alternatively, let me check the case where n=3. If there are 3 people, and we fix one person's seat, then the block can be in only one position (since the block is the entire table). Wait, but in this case, with n=3, fixing Pierre's seat, the block is the entire table. Then, the other two can be arranged in 2! ways, so total arrangements would be 2! = 2, which is correct because circular permutations of 3 is 2. So in that case, with n=3, the formula would give 2 * 2! * (something). Wait, no. Wait, in this case, the number of remaining people is 0, so it would be 2 * 2! * 0! = 2 * 2 * 1 = 4? Which is wrong. Wait, so maybe the formula is different.

Wait, in the case where n=3 and we fix Pierre's position, the block is the entire table, so there's only one position for the block (since it's the whole table). But earlier, I thought there were two positions, but that was in the case of n=8. Wait, maybe my previous logic was wrong.

Wait, let me re-examine. For n=8, fixing Pierre's seat. The block of three (Pierre, Rosa, Thomas) has to be consecutive. So, how many ways can that block be placed?

Since the table is circular, fixing Pierre at seat 1, the block can start at seat 1 and go clockwise to seats 1,2,3 or start at seat 1 and go counter-clockwise to seats 1,8,7. Therefore, two possible blocks. However, if we have a block of three seats, starting at seat 1, going clockwise, or starting at seat 1 going counter-clockwise. Therefore, two positions. However, if n=3, fixing Pierre at seat 1, the block would be seats 1,2,3 which is the entire table, but there's only one way to do that. So why in n=3, the number of positions is 1? Because if n=3, the block is the entire table, so there's only one possible block. Therefore, in general, for a block of size k in a circular table with n seats, fixing one person's position, the number of positions for the block is k? Or is it something else?

Wait, actually, if you fix one person's position, then the number of ways to place a block of size k that includes that person is equal to the number of ways to choose the rest of the block's positions. Since the block must include the fixed person, and be consecutive.

In a circular table with n seats, if we fix seat 1 as Pierre, then the block can be placed in two directions: clockwise or counter-clockwise. But if the block has size k, then depending on k, the number of positions may vary.

Wait, actually, given a fixed person, the number of ways to place a consecutive block of size k that includes that person is equal to 2 if the block is not the entire table. Wait, no. If the block is size k, then once you fix one person, the block can be placed in k positions? Wait, not sure.

Alternatively, think of the number of positions where a block of size k can be placed such that it includes the fixed person. For example, in a circular table with n seats, fixing seat 1. The number of blocks of size k that include seat 1 is equal to the number of ways to choose the other k-1 seats adjacent to seat 1 in one direction or the other. Wait, but if k is larger than 2, it's not just two.

Wait, let me take an example. If the block size is 3, and we fix seat 1, then the block can be seats 1,2,3 or seats 1,8,7 (assuming 8 seats). Therefore, two blocks. Similarly, if the block size is 4, fixing seat 1, the blocks would be 1,2,3,4 or 1,8,7,6. So two blocks. Wait, so for a block size of k, fixing one seat, the number of blocks that include that seat is 2? No, if k = n, then there's only 1 block. For k < n, it's two blocks? Wait, for example, n=8, k=3: 2 blocks. For n=8, k=2: the blocks that include seat 1 are seats 1,2 and seats 1,8. So two blocks. Similarly, for k=1, it's just seat 1. So in general, for a block size k, fixing seat 1, the number of blocks including seat 1 is 2 if k ≥ 2. Wait, but for k=2, yes, two blocks. For k=3, two blocks. Wait, but if the block size is k, and we fix one seat, the number of blocks is 2, unless k=1 or k=n. So in our problem, k=3, n=8. Therefore, fixing Pierre at seat 1, there are two blocks that include seat 1 and are of size 3. Therefore, two possible positions for the block. Then, for each of these blocks, the people inside can be arranged in different ways. However, since Pierre is fixed in seat 1, the rest of the block (Rosa and Thomas) can be arranged in 2! ways. So total arrangements for the block: 2 (positions) * 2! (internal arrangements) = 4.

Then, once the block is placed, the remaining 5 people can be arranged in 5! ways. Therefore, total number of arrangements is 4 * 5! = 4 * 120 = 480.

But earlier, the formula of treating the block as a single entity gave 5! * 3! = 720. Which is different. So which is correct?

Let me check with n=3 again. If we use this second approach: fixing Pierre's seat, there is 1 block (the whole table). The number of positions: 1. Then, internal arrangements: since the block is the entire table, and Pierre is fixed, the other two can be arranged in 2! ways. Then remaining people: zero, so 0! = 1. Therefore, total arrangements: 1 * 2! * 1 = 2, which is correct. So for n=3, the formula works. Then for n=8, the formula would be 2 * 2! * 5! = 2*2*120=480. But according to the first approach, treating the block as a single entity gave 5! * 3! = 720. Which one is correct?

Wait, let me think of another way. Suppose we don't fix Pierre's position. Let's try using the standard formula for circular permutations with a block.

In circular permutations, the formula when you have a block is (number of blocks - 1)! * internal permutations. But when the block is a single block, then number of blocks is 1. So (1 - 1)! * 3! = 1 * 6 = 6. Which is wrong, as discussed earlier. So that formula is wrong.

But when we fix one position, we get a different number. So why does fixing a position give a different answer?

Alternatively, maybe the correct formula is (number of entities - 1)! * internal permutations, but where entities are the blocks and the other individuals. However, if we don't fix positions, then we have to account for rotations. Wait, but in the case of fixing Pierre's position, we already accounted for rotational symmetry. So perhaps that approach is correct?

Alternatively, let me use another approach. The total number of circular permutations where the three people are together can be calculated by considering the following:

First, treat the three people as a single entity, so we have 6 entities (the block and the other 5 people). The number of circular permutations of these 6 entities is (6 - 1)! = 120. Then, multiply by the number of ways to arrange the three people within the block, which is 3! = 6. Therefore, total arrangements is 120 * 6 = 720. However, earlier when fixing Pierre's position, we got 480. Which is conflicting. What's going on?

Wait, perhaps when we fix Pierre's position, we are imposing an extra condition, so we get a different number. Wait, but both methods should give the same result. Therefore, there must be a mistake in one of the methods.

Wait, let's think of it in terms of linear arrangements. For linear arrangements, the number of ways where Pierre, Rosa, and Thomas are together is 6! * 3! (treating them as a block, so 6 entities, which can be arranged in 6! ways, multiplied by 3! for the block). But for circular arrangements, it's different.

But for circular arrangements, there's a formula that is similar to linear arrangements divided by n (because of rotations). Wait, but not sure.

Alternatively, consider that in circular arrangements, the number of arrangements where the three are together is equal to the number of linear arrangements where the three are together, divided by n. Because each circular arrangement corresponds to n linear arrangements (rotations). But let's check:

Number of linear arrangements where the three are together is (n - k + 1)! * k! So here, n=8, k=3. So (8 - 3 + 1)! * 3! = 6! * 3! = 720 * 6 = 4320. Wait, no, 6! is 720, times 3! is 720*6=4320. Then, if we divide by 8 (number of rotations), we get 4320 / 8 = 540. So that would be the number of circular arrangements.

But this contradicts both previous results of 480 and 720. Which one is correct?

Alternatively, maybe it's not divided by n, but something else. Wait, the formula for circular permutations where certain people are together.

Wait, another way: Let me think of it as arranging the block and the other people. In circular permutations, if we consider the block as a single entity, then we have 6 entities. The number of circular permutations is (6 - 1)! = 120. Then, multiply by 3! for the internal arrangements. So 120 * 6 = 720. However, this seems to be different from the linear approach divided by n.

But according to linear approach, we have 6! * 3! = 720 * 6 = 4320 linear arrangements. Then, if we consider circular arrangements, each circular arrangement corresponds to 8 linear arrangements (due to rotations), so number of circular arrangements would be 4320 / 8 = 540. But this contradicts with the previous result of 720.

So now, which is correct?

Alternatively, let's check with n=3 again. If n=3, the number of circular arrangements where all three are together is 2 (as before). Using the formula (n - k + 1)! * k! divided by n: (3 - 3 + 1)! * 3! / 3 = 1! * 6 / 3 = 6 / 3 = 2. Correct. Whereas the other approach, treating as block: (number of entities -1)! * internal permutations. Here, number of entities is 1, so (1 -1)! * 3! = 1 * 6 = 6. Which is wrong. Therefore, the formula of (number of entities -1)! * internal permutations is wrong? But in the linear case, the formula (n - k + 1)! * k! is correct. So perhaps for circular arrangements, the correct formula is [(number of entities -1)! * internal permutations]?

But for n=3, number of entities is 1, so [(1 - 1)! * 3!] = 6, which is wrong, but according to the division approach, we get correct result 2. Therefore, maybe the correct formula is [(number of entities -1)! * internal permutations] / something?

Wait, no. Wait, maybe the formula is different. Let me think again. In the case of circular permutations, when we have a block, we can think of it as follows:

If we have a block and other entities, the number of circular arrangements is equal to the number of linear arrangements of the entities divided by the number of rotations. Wait, but the number of linear arrangements is (number of entities)! * internal permutations. Then, to convert to circular arrangements, divide by the number of entities? Wait, no, because circular arrangements are (number of entities - 1)!.

Wait, perhaps the formula is (number of entities - 1)! * internal permutations, but this gives for n=3, 1 * 6 = 6, but actual answer is 2. Hence discrepancy.

Wait, so where is the mistake?

In the case of n=3, the formula (number of entities - 1)! * internal permutations is (1 -1)! * 3! = 6, but the correct number is 2. Therefore, the formula is wrong. Therefore, there must be another way.

Wait, but in linear arrangements, treating the block as a single entity, you get (number of entities)! * internal permutations. For n=3, that would be 1! * 3! = 6, which is the same as linear arrangements. But for circular arrangements, it's different. So how do they differ?

For circular arrangements, if we have a block, maybe the number is (number of entities - 1)! * internal permutations / something? Or is it different?

Wait, actually, in circular permutations, sometimes when you have symmetries, you have to adjust. For example, when considering necklaces with beads, sometimes you divide by symmetries.

But perhaps the standard formula is as follows:

When arranging objects around a circular table, if there are m objects, it's (m - 1)!.

If we have a block, and other objects, then the number of arrangements is (m - 1)! * (internal permutations of the block).

But in the case where the block is the only object, m=1, so (1 - 1)! * 3! = 1 * 6 = 6, but the actual number of circular arrangements is 2. Therefore, the formula is not correct in that case.

Wait, so what's different?

Ah! When we have a block, the internal permutations might be overcounted if the block has rotational symmetry. Wait, but in the case of n=3, the internal permutations are 3! = 6, but actual unique circular arrangements are 2. So perhaps the formula needs to divide by the number of rotations? But that seems complicated.

Alternatively, maybe when we treat the block as a single entity, we are overcounting because rotating the block around the table might lead to equivalent arrangements?

Wait, no, because the block is treated as an entity, and the rest of the table is arranged around it. Wait, this is getting too tangled.

Let me try to use the two methods I had earlier and see if I can find which is correct.

First method: fixing Pierre's position.

If Pierre is fixed, number of ways to arrange the rest is 2 * 2! * 5! = 4 * 120 = 480.

Second method: treating the block as a single entity, number of arrangements is (6 - 1)! * 3! = 120 * 6 = 720.

Third method: using linear arrangements divided by n: 6! * 3! / 8 = 720 * 6 / 8 = 540.

So three different answers. Which one is correct?

Let me think of a small example where I can compute manually. Let's take n=4, and want to find the number of circular arrangements where two specific people sit together.

Let's say people are A, B, C, D. Want A and B together. How many circular arrangements?

Total circular permutations for 4 people: 3! = 6.

Number of arrangements where A and B are together: ?

In circular table, fixing A's position. Then B can be to A's left or right: 2 possibilities. The other two people can be arranged in 2! ways. Total: 2 * 2 = 4. Which is equal to (number of entities - 1)! * internal permutations. Here, treating A and B as a block: number of entities is 3 (block + C + D). Then (3 - 1)! * 2! = 2! * 2 = 4. Which matches. So in this case, the formula works.

But if we use the alternative method of linear arrangements divided by n: linear arrangements where A and B are together: (n - k + 1)! * k! = (4 - 2 + 1)! * 2! = 3! * 2 = 6 * 2 = 12. Divided by 4: 12 / 4 = 3. Which is different from 4. So discrepancy here.

But manual calculation shows it's 4. Therefore, the correct formula is the first one: (number of entities - 1)! * internal permutations. So why the discrepancy with the third method?

Ah, because in the third method, when we calculate linear arrangements where the block is together, it's not just (n - k + 1)! * k! but actually (n - k + 1)! * k! ?

Wait, for n=4, k=2. Number of linear arrangements where A and B are together: treat them as a block, so 3 entities: block, C, D. These can be arranged in 3! ways, and the block can be in 2! ways, total 3! * 2! = 12. However, in circular arrangements, if we divide by n=4, we get 3, which is wrong. But manual calculation shows it should be 4. Therefore, the third method is flawed.

So why the difference? Because when converting from linear to circular, dividing by n is not always correct? Or maybe there's an adjustment needed?

Alternatively, in circular arrangements, the number of arrangements where the block is together is equal to (number of entities - 1)! * internal permutations. Which in this case works. For n=4, entities = 3, so (3-1)! * 2! = 2 * 2 = 4. Which is correct.

Therefore, maybe the formula is correct, and the third method was wrong. So why did that happen?

Because when I divided by n, I assume that each circular arrangement corresponds to n linear arrangements, but perhaps when the block is considered, some symmetries are counted differently.

Therefore, trusting the formula (number of entities -1)! * internal permutations, which worked for n=4 and n=3 (when I adjust the numbers), let me go back to the original problem.

Original problem: 8 people, with a block of 3 (Pierre, Rosa, Thomas). So number of entities: 8 - 3 + 1 = 6. Therefore, number of circular arrangements is (6 - 1)! * 3! = 5! * 6 = 120 * 6 = 720.

But earlier, when fixing Pierre's position, I got 480. Why is there a discrepancy?

Wait, if I use the formula for fixing Pierre's position: fixed Pierre, then the number of ways is 2 * 2! * 5! = 480. But according to the entity method, it's 720. Which is different.

Wait, but if I don't fix Pierre's position, the total number should be 720. But when I fix Pierre's position, I should get 720 / 8 = 90? Wait, no. Wait, fixing one position in circular permutations reduces the count by a factor of n. Wait, no. If total number is 720, then fixing one person's position would result in 720 / 8 = 90. But according to the other method, fixing Pierre's position, I have 480. Which is way bigger than 90. So something's wrong.

Wait, clearly there is a miscalculation in the fixing position method. Let me re-examine that.

If we fix Pierre's position, then we are effectively converting the circular problem into a linear one, with Pierre's position fixed. So the number of arrangements should be equal to the total number divided by n (the number of seats), but in this case, it's not matching.

Let me redo the fixing method. Let's fix Pierre at seat 1. Now, we need to arrange the remaining 7 people around the table with the condition that Rosa and Thomas are next to Pierre or to each other in a block.

Wait, no, the condition is that all three (Pierre, Rosa, Thomas) must sit together. Since Pierre is fixed at seat 1, the block of three must include seat 1. Therefore, the block can be either seats 1,2,3 or seats 1,8,7. Therefore, two possible positions for the block.

For each of these positions, the block contains Pierre, plus two others. Since Pierre is fixed, the other two can be arranged in 2! ways. So for each block position, 2! arrangements. So total for blocks: 2 * 2! = 4.

Then, the remaining 5 people can be arranged in the remaining seats. How many seats are left? Total seats are 8. The block takes 3 seats, so remaining seats: 8 - 3 = 5. So those 5 people can be arranged in 5! ways. Therefore, total arrangements with Pierre fixed: 4 * 5! = 4 * 120 = 480.

But if the total number of arrangements without fixing is 720, then fixing Pierre's position should give 720 / 8 = 90. But here, we get 480. Which is not matching. So there's a contradiction.

Wait, this suggests that my first method using entities is wrong, and the fixing method is correct? Or vice versa?

Wait, perhaps my fixing method is wrong. Let me think again. If I fix Pierre's position, the total number of arrangements where the block is together should be equal to total number of circular arrangements with the block together divided by 8 (since fixing one position accounts for rotational symmetry). So if total number is X, then fixing Pierre's position would give X / 8. But according to fixing method, I get 480, so X would be 480 * 8 = 3840, which is way higher than the entity method result of 720.

But earlier with n=4, fixing a position worked. Let me check with n=4, k=2. Total number of circular arrangements where two people are together: 4. If I fix one person's position, say A, then number of arrangements is 2 * 1! * 2! = 2 * 1 * 2 = 4. Wait, no, let me redo it.

For n=4, people A, B, C, D. Fix A at seat 1. Block of A and B must be together. The block can be seats 1,2 or seats 1,4. For each, B can be at 2 or 4. Since the block includes A and B, and A is fixed at seat 1, then B can be at 2 or 4. So two possibilities. Then the remaining two people (C and D) can be arranged in seats 3 and the remaining seat. Wait, if block is seats 1,2, then remaining seats are 3 and 4. But seat 4 is adjacent to seat 1. Wait, if block is seats 1,2, then seats 3 and 4 are remaining. So C and D can be arranged in 2! ways. Similarly, if block is seats 1,4, then remaining seats are 2 and 3, which can be arranged in 2! ways. So total arrangements: 2 (block positions) * 2! (arrangements of remaining) = 2 * 2 = 4, which matches the total circular arrangements (which is 4). So in this case, fixing position gives the correct result, and the total number is 4. If I had used the entity method: number of entities = 4 - 2 + 1 = 3. (3 - 1)! * 2! = 2! * 2 = 4. Correct.

So why in the original problem, fixing Pierre's position gives 480, while the entity method gives 720? Let's check if there's an error in the fixing method. For n=8, fixing Pierre's position, number of arrangements where the block is together is 4 * 5! = 480. If we trust this, then total number of circular arrangements should be 480 * 8 = 3840. But according to the entity method, it's 720. Which is different.

But for n=4, the entity method and fixing position method gave the same result. So why in n=8, they differ? There must be a miscalculation in one of the methods.

Wait, in the n=4 example, the number of entities was 3, leading to (3-1)! * internal permutations = 2! * 2 = 4, which matches the fixing method. So in that case, the total number was 4, and with fixing position, also 4. So why for n=8, fixing gives 480, while the entity method gives 720?

Wait, let me recast the entity method for n=8. Number of entities is 6 (block + 5 people). Circular arrangements: (6-1)! * 3! = 120 * 6 = 720. So total arrangements is 720. If I fix Pierre's position, the number should be 720 / 8 = 90. But according to fixing method, it's 480. Therefore, there's a contradiction.

Therefore, one of the methods must be wrong. Let me try to find the error.

Wait, when I fixed Pierre's position, I considered that the block could be in two positions (seats 1-2-3 or 1-8-7). But maybe there are more positions?

Wait, no. With n=8 and block size 3, fixing Pierre at seat 1, the block must include seat 1 and be consecutive. The only possible blocks are seats 1-2-3 and seats 1-8-7. So two blocks. So why is there a discrepancy?

Wait, maybe the error is that when I fixed Pierre's position, I didn't account for the fact that the block can be rotated? Or is there a miscalculation in the internal arrangements?

Wait, if the block is in seats 1-2-3, then seat 1 is Pierre, and seats 2 and 3 are Rosa and Thomas. They can be arranged in 2! ways. Similarly, if the block is in seats 1-8-7, seat 1 is Pierre, seats 8 and 7 are Rosa and Thomas, also 2! ways. So total for the block: 2 * 2! = 4. Then remaining 5 people: seats 4,5,6,7,8? Wait, no. If the block is seats 1-2-3, then remaining seats are 4,5,6,7,8. But seat 8 is adjacent to seat 1. Wait, no, if the block is seats 1-2-3, then the remaining seats are 4,5,6,7,8. Wait, but seat 8 is separate. Yes, so those 5 seats can be filled with the remaining 5 people in 5! ways. Similarly, if the block is seats 1-8-7, remaining seats are 2,3,4,5,6. Which are also 5 seats. So 5! ways. So total is 2 * 2! * 5! = 4 * 120 = 480.

But according to entity method, it should be 720. So why the difference? Let me think if there is another way to place the block.

Wait, if we don't fix Pierre's position, but consider all possible positions for the block, how many are there? For a block of size 3 in a circular table of 8 seats, the number of blocks is 8 (each starting at a different seat). But since rotations are considered the same, if we don't fix any position, the number is 8 / 8 = 1? No, that's not right.

Actually, the number of distinct blocks in circular table is n (number of seats), but since rotations are considered the same, but in our case we are counting all possible blocks (which would be n, but considering rotations as distinct? No, in the entity method, we treat the blocks as distinct entities.

Wait, this is getting too confusing. Maybe I should trust the fixing method, since it worked for n=4 and it seems logical.

Alternatively, think of the entity method. If we have 6 entities (block + 5 people), then number of circular arrangements is (6-1)! * 3! = 120 * 6 = 720. But if we fix one position, say, the block's position, then we have to adjust.

Wait, let me think: if we fix the block's position, then the number of arrangements is 3! * 5! (since block is fixed, internal arrangements are 3! and remaining people are 5!). But since we fixed the block's position, we don't have to divide by anything. But how many ways to fix the block's position?

Wait, in circular table, fixing the block's position would mean that we can rotate the table so that the block is in a specific spot. So if we fix the block at a specific position, then the number of arrangements is 3! * 5! = 6 * 120 = 720. But since there are 8 seats, and the block takes up 3, the number of positions for the block is 8, but since rotations are considered the same, fixing the block's position accounts for all rotations. Wait, no, fixing the block's position means we're considering rotational symmetry already. Therefore, the total number of circular arrangements is 3! * 5! = 720. This matches the entity method.

But earlier, when I fixed Pierre's position, I got 480. Why? Because fixing Pierre's position is a different constraint. If instead of fixing the block's position, I fix Pierre's position, it affects the count. Let me see.

If I fix the block's position, then Pierre can be in any position within the block. But if I fix Pierre's position, it restricts the block's position. Hence, the different counts.

Therefore, the correct answer is 720, from the entity method. But why does fixing Pierre's position give 480? Probably because by fixing Pierre's position, I'm not accounting for all possible rotations where the block is placed relative to Pierre.

Wait, but in fixing Pierre's position, I considered all possible placements of the block that include Pierre. So why is it different?

Let me think: total number of circular arrangements is 720. If I fix Pierre's position, how many arrangements are there? It should be 720 / 8 = 90. But according to my calculation, it's 480. So there is a miscalculation.

Wait, let me recast the fixing Pierre's position method. Suppose we fix Pierre at a seat. The total number of arrangements where the three are together is equal to the number of ways to arrange the block and the others.

But if I fix Pierre's position, then the block is determined by where Rosa and Thomas are. But how many ways are there?

Alternatively, think of it as: once Pierre is fixed, the block of three can be arranged in 2 * 2! ways (as before), and the rest arranged in 5! ways, giving 4 * 120 = 480. But total number of arrangements with Pierre fixed is 480. But total number of circular arrangements is 720. So if I fix Pierre's position, the number should be 720 / 8 = 90, but it's 480. So clearly, something is wrong.

Wait, but if I fix Pierre's position, the total number of arrangements without any restrictions is (8-1)! = 5040? No, wait. If we fix Pierre's position, then the number of circular arrangements is (8-1)! / 8 = 7! / 8 = 5040 / 8 = 630? No, wait: when you fix one person's position in a circular arrangement, the number of distinct arrangements is (n-1)! So for 8 people, fixing one person, it's 7! = 5040? No, no! Wait, if you fix one person's position, the number of arrangements is (n-1)! / n * n = (n-1)! Wait, no. Let me clarify.

When you fix one person's position in a circular arrangement, you're effectively converting it into a linear arrangement. The number of distinct circular arrangements is (n-1)! So if you fix one person's position, the number of linear arrangements is (n-1)! So for n=8, fixing Pierre's position, the number of arrangements is 7! = 5040? No, that can't be. Wait, no. Wait, for n people, the number of circular arrangements is (n-1)! If you fix one person's position, you're removing the rotational symmetry, so the number of arrangements is (n-1)! / n * n = (n-1)!? Wait, I'm confused.

Actually, fixing one person's position in a circular table makes it equivalent to arranging the remaining (n-1) people linearly. So the number of arrangements is (n-1)!.

Wait, for n=8, fixing Pierre's seat, the number of arrangements is 7! = 5040. But earlier, when I calculated the number of arrangements where the block is together, I found 480. So 480 / 5040 = 480 / 5040 = 4/42 = 2/21. Which is the probability that the three are together, which seems plausible?

But if the total number of arrangements is 720, and the number of arrangements with block together is 720, then fixing Pierre's position would give 720 / 8 = 90. But according to fixing method, it's 480. So clearly, there is a miscalculation.

Wait, let me check what is the total number of arrangements with the block together. If it's 720, and total number of circular arrangements is 7! = 5040, then the probability is 720 / 5040 = 1/7. But if I fix Pierre's position, the probability that the block is together is 480 / 7! ? 480 / 5040 = 4/42 = 2/21 ≈ 0.095, while 1/7 ≈ 0.142. So they don't match. So one must be wrong.

Alternatively, let me calculate the total number of circular arrangements where the three are together correctly.

Using the entity method: 5! * 3! = 720. Total circular arrangements: 7! = 5040. So the probability is 720 / 5040 = 1/7 ≈ 0.142.

Now, using the fixing method: fixing Pierre's position, number of such arrangements is 480, and total arrangements with Pierre fixed is 7! = 5040? No, wait: if we fix Pierre's position, the total number of arrangements is (8-1)! = 5040? No, no! Wait, if you fix one person's position in a circular table, the number of distinct arrangements is (n-1)! So for n=8, fixing Pierre's seat, it's 7! = 5040? Wait, no! Wait, no. Wait, when you fix one person's position, you are effectively making the problem linear. The number of arrangements is (n-1)! So for n=8, it's 7! = 5040? No! Wait, no. Wait, the number of circular permutations is (n-1)! So if you fix one person, the number of arrangements is (n-1)!

Wait, for example, for n=3, fixing one person, the number of arrangements is 2! = 2, which is correct. For n=4, fixing one person, it's 3! = 6, which is correct. So yes, for n=8, fixing one person, the number of arrangements is 7! = 5040. But earlier, I thought it was (n-1)! = 5040, which is correct.

But according to fixing method, the number of arrangements with the block together is 480, so the probability is 480 / 5040 = 4/42 = 2/21 ≈ 0.095, while the correct probability is 1/7 ≈ 0.142. Hence, discrepancy.

Therefore, there must be an error in the fixing method. Let me find it.

When I fixed Pierre's position, I assumed that the block could be in two positions (seats 1-2-3 or 1-8-7), each with 2! internal arrangements, and the rest arranged in 5! ways. But perhaps this is missing some arrangements?

Wait, maybe there are more ways to position the block? Or maybe the internal arrangements are not just 2!?

Wait, if the block is fixed to include Pierre, but the block can be in more positions? No, for n=8, block size 3, fixing Pierre at seat 1, there are only two possible blocks: 1-2-3 and 1-8-7. So that seems correct.

Wait, but perhaps the internal arrangements are not just 2!?

Wait, in each block, Pierre is fixed at seat 1. The other two seats in the block are occupied by Rosa and Thomas. So for the block 1-2-3, seats 2 and 3 are Rosa and Thomas, which can be arranged in 2! ways. Similarly, for block 1-8-7, seats 8 and 7 are Rosa and Thomas, which can be arranged in 2! ways. So total of 2 * 2! = 4. Then the rest of the seats, 5 people, arranged in 5! ways. So total 4 * 120 = 480.

But if the correct total number is 720, then where is the missing 720 - 480 = 240?

Ah! Wait, perhaps the block can be in more positions? For example, if the block is not required to include Pierre's seat? Wait, no! The condition is that Pierre, Rosa, and Thomas all sit together. Since Pierre is fixed, the block must include Pierre. So the block must include seat 1.

Wait, but what if the block is somewhere else, but still includes Pierre? No, because Pierre is fixed at seat 1. So the block must include seat 1. Therefore, only two possible blocks.

Wait, but what if the block is placed such that Pierre is in the middle of the block? For example, if the block is seats 8, 1, 2. Which is the same as seats 1-2-8, but that is already counted as block 1-8-7 or 1-2-3? No. Wait, block 1-8-7 is seats 1, 8, 7. Block 1-2-3 is seats 1, 2, 3. So if the block is seats 8, 1, 2, that's the same as block 1-2-8, which is the same as block 1-2-3? No, because in a circular table, the block seats 1, 2, 3 is consecutive, and block 1, 8, 7 is also consecutive. But block 1, 2, 8 is not consecutive, since 2 and 8 are not adjacent. So no, that's not a valid block.

Therefore, only two possible blocks that include seat 1. So why is there a discrepancy?

Wait, let me think differently. Let me calculate the number of ways where Pierre, Rosa, and Thomas are together, with Pierre fixed at seat 1.

The block of three can start at seat 1 and extend clockwise or counter-clockwise. For each direction, the block is three seats. So, for each direction, the block is determined, and within the block, Pierre is fixed, and the other two can be arranged. But perhaps there are more ways to arrange the block?

Wait, if the block is allowed to be in any position, but includes Pierre, then there are more positions? No, because Pierre is fixed, and the block must be consecutive. So only two positions.

Wait, here's a thought: perhaps the block can include Pierre in different positions. For example, if the block is three seats, and Pierre is in the middle or at the end. But since we're fixing Pierre at seat 1, he can only be at the start of the block, or in the middle? Wait, no, if the block is three seats, and Pierre is at seat 1, then he is at the start of the block if the block is 1-2-3, or at the end of the block if the block is 7-8-1. Wait, but in this case, the block is 7-8-1, which includes seat 1. So in this case, Pierre is at the end of the block. But in this case, the block is 7-8-1, which is consecutive. So, but in this case, the block can be arranged such that Pierre is at the end, and Rosa and Thomas are at 7 and 8. But this is already counted in the block positions 1-8-7. So it is already included in the two block positions.

Therefore, no additional block positions.

So why is the count different?

Wait, let me think about the number of ways to have Pierre, Rosa, Thomas together with Pierre fixed. Let's think of it as: the number of ways where Pierre is at seat 1, and Rosa and Thomas are adjacent to him or to each other.

Wait, but they must all be together. So the three of them must occupy three consecutive seats, with Pierre at seat 1. So the three seats must be 1-2-3 or 1-8-7. In each of these cases, the other two people (Rosa and Thomas) can be arranged in 2! ways. Then, the remaining 5 people can be arranged in 5! ways. So total 2 * 2! * 5! = 480.

But if the correct answer is 720, then where is the error?

Wait, perhaps fixing Pierre's position is not the right approach here, and the entity method is correct? Or maybe the entity method overcounts?

Wait, let me think of it this way: in the entity method, we treat the block as a single entity, giving us 6 entities. Number of circular arrangements is (6-1)! * 3! = 120 * 6 = 720. But if we calculate it by fixing one person, we get 480.

But if I don't fix any positions, and calculate using the entity method, it should be correct. Therefore, there must be a mistake in the fixing method.

Let me try a different approach: let's calculate the number of circular arrangements where Pierre, Rosa, and Thomas are together by considering all possible positions of the block.

In a circular table with 8 seats, the number of ways to choose a block of 3 consecutive seats is 8 (since it's circular). For each block, we can arrange Pierre, Rosa, Thomas in 3! ways. Then, arrange the remaining 5 people in the remaining 5 seats in 5! ways.

However, in circular arrangements, this counts each arrangement multiple times. Specifically, each arrangement is counted once for each rotational position of the block. Wait, but since the block is fixed in position, and we are counting all blocks, this may not account for rotational symmetry.

Wait, no. If we calculate it as 8 (number of blocks) * 3! * 5! = 8 * 6 * 120 = 5760. But this is the number of linear arrangements, because each block is considered distinct, and rotations are not accounted for. To convert to circular arrangements, we need to divide by the number of rotations, which is 8. So 5760 / 8 = 720. Which matches the entity method!

Ah! So this is the correct way. So the error in the fixing method was that I fixed Pierre's position and didn't consider all possible blocks. By considering all blocks and dividing by the number of rotations, we get 720.

Therefore, the correct answer is 720. So why did the fixing method give 480? Because when I fixed Pierre's position, I restricted the block to only those that include Pierre, but in reality, the block can be anywhere, and fixing Pierre's position still allows for other blocks that include Pierre. Wait, no. Wait, in this calculation, when we fix Pierre's position, we are not considering all possible blocks, but only those that include Pierre. But if we want to calculate the total number of arrangements where the block includes Pierre, we should calculate it as follows: number of blocks including Pierre is 8 (total blocks) * ... wait, no.

Actually, the number of blocks that include Pierre is equal to the number of blocks where Pierre is in the block. Since each block of 3 consecutive seats contains 3 people, and there are 8 blocks, the total number of blocks is 8. How many of these blocks include Pierre? For each block, it can start at any seat. For example, block starting at seat 1: includes seats 1-2-3. Block starting at seat 2: seats 2-3-4. ... Block starting at seat 8: seats 8-1-2. So, how many of these blocks include seat 1? Blocks starting at seat 8 (includes 8-1-2), seat 1 (1-2-3), and seat 2 (2-3-4) does not include 1. Wait, no. Block starting at seat 8 includes seat 8, 1, 2. Block starting at seat 1 includes 1, 2, 3. Block starting at seat 8 and 1 include seat 1. So total of 2 blocks that include seat 1. Therefore, out of 8 total blocks, 2 include seat 1. So if we fix Pierre's position at seat 1, the number of blocks that include him is 2. So, for each of these 2 blocks, we can arrange the three people in 3! ways, and the rest in 5! ways. So total is 2 * 3! * 5! = 2 * 6 * 120 = 1440. But this is the number of arrangements where the block includes Pierre, but since we fixed Pierre's position, these are the number of arrangements where the block includes Pierre. But since in total, there are 8 blocks, each corresponding to a different position, but fixed Pierre's position restricts it.

Wait, but this seems to give 1440, which is higher than the total number of circular arrangements (5040). But no, because when we fix Pierre's position, the total number of arrangements is 7! = 5040, and the number of arrangements where the block includes Pierre is 1440, which is 1440 / 5040 = 1440 / 5040 = 12/42 = 2/7, which is different from the earlier probability of 1/7. Wait, but this is contradictory.

This shows that there is confusion in the fixing method. It seems that the correct answer is 720, as per the entity method and the block counting divided by rotations. Therefore, I must have made a mistake in the fixing method.

Given that, and since the entity method and the block counting method both give 720, I will trust that the correct answer is 720.

So, the number of ways is 5! * 3! = 120 * 6 = 720.

Therefore, the final answer is \boxed{720}.

**Final Answer**
\boxed{720}

To determine how many ways 8 people can sit around a round table **such that** Pierre, Rosa, and Thomas all sit **together**, we proceed by considering circular permutations with the specific constraint that a particular group (of size 3) must **be seated consecutively**.

---

### Step 1: Understand Circular Permutations with Group Constraints

In standard **circular permutations**, the number of distinct arrangements of $ n $ people is $ (n-1)! $. This is because **rotations** of the table are considered **identical**, so we fix one person’s position to eliminate overcounting.

However, when a **group** of people must all sit **together**, we can treat this group as a **single "block"**. This effectively reduces the problem to arranging this block with the remaining individuals.

---

### Step 2: Break the Problem into Two Parts

- **Group as a block**:
  We consider the 3 people (Pierre, Rosa, Thomas) as a block, which can be arranged internally in $ 3! $ ways.

- **Remaining people**:
  The remaining 5 people plus the "block" form $ 6 $ entities to be arranged in a circle.

---

### Step 3: Count the Circular Arrangements of the 6 Entities

Since we're dealing with **circular permutations** of these 6 entities, the number of arrangements is:

$$
(6 - 1)! = 5! = 120
$$

---

### Step 4: Multiply by Internal Arrangements of the Block

The block of 3 people can be arranged among themselves in $ 3! = 6 $ ways.

---

### Step 5: Final Calculation

$$
\text{Total number of valid arrangements} = 5! \times 3! = 120 \times 6 = \boxed{720}
$$

---

### Final Answer

$$
\boxed{720}
$$

Solution: First choose three consecutive seats for Pierre, Rosa, and Thomas.  It doesn't matter which three consecutive seats that we choose, since any three such seats can be rotated to any other such seats.  Once the three seats are chosen, there are $3!$ ways to seat the three friends there.  The other five seats are for the other five people, so there are $5!$ ways to seat them there.  The answer is $3! \times 5! = \boxed{720}$.