Okay, so I need to figure out how many ways 8 people can sit around a round table with the condition that Pierre, Rosa, and Thomas all want to sit together. Hmm, circular permutations can be a bit tricky, but let me try to break it down step by step.

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

If Pierre, Rosa, and Thomas must sit together, maybe I can treat them as a single unit or block. Let's see. If we group them together, then instead of 8 individuals, we have this block plus the other 5 people. So that would be 6 units in total. But wait, since it's a circular table, the number of arrangements would be (6-1)! = 5! for the blocks. But also, the three people within the block can be arranged among themselves in different ways. How many ways can Pierre, Rosa, and Thomas be arranged within their block?

Well, for three people, the number of permutations is 3!. So multiplying that by 5! should give the total number of arrangements. Let me check that again. So total arrangements would be 5! * 3!.

But wait, is there any other consideration here? For circular tables, sometimes people fix one position to account for rotations. But since we already considered the blocks as a single unit and calculated (6-1)! for circular permutations, do we need to adjust anything else?

Wait, maybe not. Because when we treat the block as a single element, the circular arrangement formula (n-1)! already accounts for the rotational symmetry. So if we have 6 units, it's (6-1)! = 5! for the blocks. Then multiply by the permutations within the block, which is 3!. So the total should be 5! * 3!.

Let me compute that. 5! is 120, and 3! is 6. So 120 * 6 = 720. So 720 ways. But wait, let me verify this with another approach to make sure I didn't make a mistake.

Alternatively, maybe I can fix the position of the block to eliminate rotational symmetry. If I fix Pierre, Rosa, and Thomas in a specific order around the table, then the remaining 5 people can be arranged in the remaining seats. But since the block can be rotated among themselves, I need to consider all possible orders within the block. Wait, but if I fix the block's position, then their internal arrangement is 3! and the others are 5!, which is the same as before. So that gives the same result.

Alternatively, if I don't fix the block, the number of ways to arrange the block around the table is (number of ways to arrange the block) * (number of ways to arrange the remaining people in the remaining seats). But since the table is circular, arranging the block is equivalent to arranging 6 units (block + 5 people) around the table, which is (6-1)! = 5! for the block's position. Then multiply by 3! for the internal arrangement. So again, 5! * 3! = 720.

Hmm, seems consistent. But let me think again. Suppose we have 8 seats around a table. If we group Pierre, Rosa, and Thomas into a block, we can imagine that block taking up 3 consecutive seats. The number of ways to arrange the block around the table would be 8 positions, but since it's a circular table, fixing the block's position in one place accounts for rotations. Wait, maybe another way to think about it is, how many ways to place the block in the 8 seats. Since it's circular, the number of distinct positions for the block is 8 (since rotating would make them equivalent). But actually, no, because once we fix the block's position, the rest can be arranged around it. Wait, maybe this approach complicates things.

Alternatively, if we consider the block as a single entity, we have 6 entities total (block + 5 individuals). In circular permutations, the number of ways to arrange these 6 entities is (6-1)! = 5! as before. Then multiply by 3! for the internal arrangements. So that's the same as before.

Alternatively, if we don't use the block approach, we could fix the position of one person to eliminate rotational symmetry. Let's say we fix Pierre at a specific seat. Then Rosa and Thomas must sit either to his immediate left or right, but since they all must sit together, they can form a block either to his left or right. Wait, but they could be in any order. Hmm, maybe this complicates things because now we have to consider the different possible arrangements of the block relative to Pierre.

Alternatively, if we fix Pierre's position, then the block of Rosa and Thomas can be arranged in two different ways relative to Pierre (either Rosa and Thomas to his left or right in some order). Wait, but actually, if we fix Pierre's position, then the block of Rosa and Thomas can be arranged in 2! ways (Rosa first then Thomas, or vice versa) and then the rest of the people can be arranged in 5! ways. But wait, but Pierre is fixed, so the total would be 2! * 5! = 2 * 120 = 240. But that contradicts the previous answer of 720. So which one is correct?

Wait, perhaps fixing Pierre's position is a different way of approaching the problem. Let me see. If we fix Pierre's position, then the block of Rosa and Thomas can be placed either to his left or right. But since the table is circular, fixing Pierre's position removes rotational symmetry. Then, Rosa and Thomas can be arranged in 2! ways in the block, and the remaining 5 people can be arranged in 5! ways. So total is 2! * 5! = 240. But this is different from the previous 720. So which is correct?

Wait, maybe the problem is that when we fix Pierre's position, we are not considering all possible rotations. But in circular permutations, fixing one person's position already accounts for rotational symmetry. So if we fix Pierre's position, then the number of arrangements is (number of ways to arrange the block and the rest). But if we fix Pierre, then Rosa and Thomas can be in 2! ways as a block adjacent to Pierre, but also, the block can be anywhere else around the table. Wait, no. If we fix Pierre's position, the block of Rosa and Thomas can be in any of the remaining 7 seats, but they must be consecutive. Wait, but if Pierre is fixed, then the block of Rosa and Thomas must consist of two consecutive seats adjacent to Pierre. Because otherwise, if they are not adjacent to Pierre, then Pierre wouldn't be in the block. Wait, but Pierre is part of the block? Wait, no. Wait, Pierre is one of the three people who must sit together. So the block includes Pierre, Rosa, and Thomas. So if we fix Pierre's position, then Rosa and Thomas must sit next to him in some order. So the block would consist of Pierre, Rosa, and Thomas in some order, but with Pierre fixed. Wait, but if we fix Pierre's position, then Rosa and Thomas can be arranged in 2! ways next to him on either side. Wait, but if the block is three people, then fixing Pierre's position, Rosa and Thomas can be arranged in two different orders either to his left or right. Wait, but actually, if Pierre is fixed, the block would consist of Pierre, Rosa, Thomas arranged in a line. But since it's a circular table, the block can be arranged in 3! ways, but since Pierre is fixed, the other two can be in 2! ways. But wait, if Pierre is fixed, then Rosa and Thomas can be arranged in two positions relative to him, but the block itself can be rotated around the table.

Wait, this is getting confusing. Let me approach it differently.

Method 1: Treat the three people as a block. Then, there are 6 units (block + 5 individuals). For circular permutations, the number of arrangements is (6-1)! = 120. Then, within the block, the three can be arranged in 3! = 6 ways. So total is 120 * 6 = 720.

Method 2: Fix Pierre's position. Then, the block of Rosa and Thomas must include Pierre? Wait, no. Wait, if Pierre is fixed, then the block of three (Pierre, Rosa, Thomas) must be arranged around the table. But if Pierre is fixed, then Rosa and Thomas can be arranged in two positions relative to him. Wait, but the block needs to have three people. If Pierre is fixed, then Rosa and Thomas can be either to his left or right, but in two different orders. Wait, but the block is three people, so if Pierre is fixed, then Rosa and Thomas can be either to his left or right, but in two different orders. So the block can be arranged in 2! ways (Rosa first then Thomas, or Thomas first then Rosa) in the two seats adjacent to Pierre. Then, the remaining 5 people can be arranged in the remaining 5 seats in 5! ways. So total arrangements would be 2! * 5! = 2 * 120 = 240.

But now, which method is correct? 720 vs 240. There's a discrepancy here. Where did I go wrong?

Wait, in Method 2, by fixing Pierre's position, we are effectively considering the block as part of the arrangement where Pierre is fixed. But if Pierre is fixed, then the block must include Pierre, Rosa, and Thomas. So the two seats next to Pierre must be occupied by Rosa and Thomas in some order. Then the remaining 5 people can be arranged in 5! ways. So that gives 2! * 5! = 240. But this contradicts the first method.

But why is there a difference? Because in Method 1, we are treating the entire block as a single unit, so the number of units is 6. However, when we fix Pierre's position, we are effectively reducing the problem to arranging the remaining units. But perhaps the first method is incorrect because when you treat the block as a single unit and multiply by 3!, you are considering all possible rotations of the block, but in reality, once you fix Pierre's position, the block can't rotate anymore. So maybe the correct answer is 240.

Wait, but let's think about the block. If we treat the three as a block, the number of circular arrangements is (n-1)! where n is the number of units. Here, n = 6 (block + 5 people). So (6-1)! = 120. Then multiply by the internal arrangements 3! = 6, giving 720. But in Method 2, by fixing Pierre's position, we get 2! * 5! = 240. So which is correct?

Wait, perhaps the issue is that when we fix Pierre's position, we are not accounting for all possible positions of the block. For example, the block of three could be anywhere around the table, not necessarily including Pierre. Wait, but if the three people must sit together, then the block must consist of all three. So if we fix Pierre's position, then the block must include him, but Rosa and Thomas can be arranged either to his left or right. But if we don't fix Pierre's position, then the block could be placed anywhere, and Pierre is just one of the three.

Wait, perhaps the first method is correct because it doesn't fix any positions, just treats the block as a unit. The second method is also correct, but it's only considering the cases where Pierre is fixed. But the problem states that the three people must sit together, but it doesn't specify that Pierre must be in a particular position. So perhaps the second method is missing some arrangements where the block is not including Pierre's fixed position.

Wait, no. If we fix Pierre's position, then we are considering all possible arrangements where Pierre is part of the block. But the problem states that Pierre, Rosa, and Thomas must all sit together. So their block can be anywhere around the table, not necessarily including Pierre. Wait, no, if we fix Pierre's position, then the block has to include him. So perhaps the problem is that fixing Pierre's position is too restrictive.

Wait, maybe the correct approach is to use the block method without fixing any positions. So, treating the three as a block, we have 6 units. The number of circular permutations is (6-1)! = 120. Then multiply by 3! for the internal arrangements. So 120 * 6 = 720. Therefore, the answer is 720.

But then why does fixing Pierre's position give 240? Because in that case, we are only considering arrangements where Pierre is part of the block. But if the block can be anywhere, then fixing Pierre's position is just one specific case. So actually, the total number of arrangements should be 720, which includes all possible positions of the block, not just those where Pierre is fixed. So the second method is incorrect because it only counts the cases where Pierre is in the block and is fixed, which is not the case here.

Therefore, the correct answer should be 720. But let me verify with a smaller example. Suppose we have 3 people around a table, all must sit together. Then the number of arrangements is (3-1)! = 2. But treating them as a block, we have 1 block, so (1-1)! = 1, multiplied by 3! = 6. Wait, that gives 6, which is wrong. The correct answer is 2. So something's wrong here.

Wait, that's a problem. So in the case of 3 people around a table, all sitting together, the correct number is 2 (since circular permutations of 3 people is 2). But according to the block method, it would be (1 block) so (1-1)! = 1, multiplied by 3! = 6, which is 6. That's wrong. So clearly, the block method isn't working here. Hmm. So that suggests that the block method may not always work? Or perhaps I'm misunderstanding how the block method applies to circular permutations.

Wait, maybe the issue is that when the block size equals the total number of people, which is 3 in this case, then treating them as a block results in only 1 block, leading to (1-1)! which is 1. Then multiply by 3! for internal arrangements. But in reality, when all 3 are together in a circular table, there are only 2 distinct arrangements (clockwise and counter-clockwise). So 3! / 3 = 2, which is the correct answer. So why does the block method give 6? Because when you treat the block as a single unit, you have only 1 unit, so (1-1)! = 1. Then multiply by 3! = 6. But the actual number is 2. So discrepancy here.

Therefore, the block method may not be applicable when the block size equals the total number of people? Or perhaps there's a different consideration. Wait, no, in the original problem, the block is size 3, and total people are 8. So the block is a part of the whole. But in the smaller case, when the block size is equal to the total, the block method gives a different answer. So perhaps the block method needs to be adjusted when the block size is equal to the total number of units?

Wait, maybe the formula is (n - k + 1)! / n, where n is total people and k is the size of the block? Wait, not sure. Alternatively, maybe the formula is (n - k)! * k! when considering circular permutations? But in the smaller case, that would be (3 - 3)! * 3! = 1 * 6 = 6, which is still wrong. So perhaps the block method isn't directly applicable here.

Wait, perhaps the issue is that when the block is the entire set, the circular arrangement is only 1 (since all seats are occupied by the block), so the number of arrangements is 1 * k! / n? Wait, I'm getting confused.

Alternatively, maybe the correct formula is (n - k)! * k! when the block is treated as a single unit, but in the case where the block is the entire set, n - k = 0, which would be undefined. So perhaps the formula is only applicable when k < n.

In the original problem, n = 8, k = 3, so 8 - 3 = 5. So (5)! * 3! = 120 * 6 = 720. That works. But in the smaller case, n = 3, k = 3. So (3 - 3)! * 3! = 1 * 6 = 6. But the actual answer is 2. So discrepancy again. So perhaps the formula is not universally applicable.

Alternatively, maybe the formula (n - 1)! when considering circular permutations, and then if there's a block of size k, then the formula becomes (n - k)! * k! / (n - k + 1)! or something else? Hmm, not sure.

Alternatively, maybe the correct approach is to use the formula for circular permutations with a block: (number of ways to arrange the block) * (number of ways to arrange the remaining people) divided by the number of rotations. Wait, but this is getting too vague.

Wait, perhaps the key is that when the block is treated as a single unit, the number of units is (n - k + 1). Wait, no. Wait, in the original problem, n = 8, k = 3. So the number of units is 8 - 3 + 1 = 6? Wait, 6 units: 5 individuals and 1 block. Then circular permutations of 6 units is (6 - 1)! = 120. Then multiply by 3! for internal arrangements. So 120 * 6 = 720. Which matches the original answer.

In the smaller case, n = 3, k = 3. Then number of units is 3 - 3 + 1 = 1. Then circular permutations of 1 unit is (1 - 1)! = 1. Then multiply by 3! = 6. But the correct answer is 2. So discrepancy again.

Hmm. So maybe the formula is (number of units - 1)! * (internal arrangements). Number of units is (n - k + 1). So in the original problem, 6 units, (6 - 1)! * 3! = 5! * 6 = 720. Correct.

In the smaller case, n = 3, k = 3, so 1 unit. (1 - 1)! * 3! = 1 * 6 = 6. But actual answer is 2. So discrepancy. So perhaps the formula is not applicable when the block size equals the total number of seats. Alternatively, maybe the formula is (n - k)! * k! in some cases. Wait, in the original problem, 8 - 3 = 5. So 5! * 3! = 120 * 6 = 720. In the smaller case, 3 - 3 = 0. 0! * 3! = 1 * 6 = 6. But again, wrong.

Wait, maybe the formula is (n - k)! * k! when the block is treated as a linear arrangement, but in circular permutations, we have to adjust. Wait, perhaps the correct formula is (n - k)! * k! when arranging in a line, but for circular permutations, it's (n - k - 1)! * k! ?

Wait, this is getting too convoluted. Let me think again. Let's use the standard formula for circular permutations with a block. The standard formula is (number of units - 1)! * (internal arrangements). The number of units is (n - k + 1). Wait, in the original problem, n = 8, k = 3. So units = 8 - 3 + 1 = 6. So (6 - 1)! * 3! = 5! * 6 = 720. Correct.

In the smaller case, n = 3, k = 3. So units = 3 - 3 + 1 = 1. Then (1 - 1)! * 3! = 1 * 6 = 6. But the correct answer is 2. So why is there a discrepancy here?

Wait, perhaps when the block size equals the total number of seats, the formula doesn't hold. Because in that case, there's only one unit, which is the entire circle. So the number of circular permutations is 1, but multiplied by k! gives k!, which is 3! = 6. But the actual answer is 2. So perhaps the formula is incorrect in that edge case.

Alternatively, perhaps the formula is correct, but the way we count the internal arrangements is different. Wait, in the smaller case, if all three people must sit together, then the number of circular permutations is (3 - 1)! = 2, which is correct. But according to the block formula, treating the block as a single unit, which is 1 unit, then (1 - 1)! * 3! = 1 * 6 = 6, which is wrong. So the formula overcounts here.

Therefore, maybe the formula only works when the block size is less than the total number of seats. So in the original problem, since 3 < 8, the formula works. In the smaller case, when 3 = 3, the formula is invalid. Therefore, we need a different approach when the block size equals the total number of seats. But in our original problem, 3 < 8, so the formula is applicable.

Therefore, in the original problem, the answer is 720. But let me check another example to confirm. Suppose n = 4 people, and a block of 2 people. Then according to the formula, number of units = 4 - 2 + 1 = 3. Then (3 - 1)! * 2! = 2! * 2 = 4. But let's compute manually. If we have 4 people, A, B, C, D, and A and B must sit together. How many circular arrangements?

Fix the block of A and B. They can be arranged in 2 ways (AB or BA). Then the remaining two people, C and D, and the block. So total units: 3. Number of circular arrangements: (3 - 1)! = 2. Then total arrangements: 2 * 2 = 4. Which matches the formula. So that works.

Another example: n = 4, block size 1. Then formula would be (4 - 1 + 1) = 4 units. (4 - 1)! * 1! = 6 * 1 = 6. Which is correct because each person is their own block, and circular permutations are (4 - 1)! = 6. So correct.

Therefore, seems like the formula works when the block size is less than or equal to the total number of seats, except when the block size equals the total number of seats. Wait, but in the case when block size equals total seats, like n = 3, k = 3, the formula gives 6, but actual answer is 2. So discrepancy. Therefore, perhaps the formula is:

If the block size k is equal to n, then the number of arrangements is (k - 1)! = (n - 1)!.

Else, number of arrangements is (n - k + 1 - 1)! * k! = (n - k)! * k!.

Therefore, in the original problem, n = 8, k = 3. Since k < n, the formula gives (8 - 3)! * 3! = 5! * 6 = 720. Which is correct.

In the case when k = n, like n = 3, k = 3, it's (3 - 1)! = 2, which is correct.

In another example, n = 4, k = 4. Then the number of arrangements is (4 - 1)! = 6. Which is correct.

Therefore, the formula is:

If k = n, then answer is (n - 1)!.

Else, answer is (n - k)! * k!.

Therefore, in the original problem, since k = 3 < 8 = n, the answer is (8 - 3)! * 3! = 5! * 6 = 720. So that's the correct answer.

Therefore, the final answer is 720.
</think>