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. And rotations are considered the same seating. Hmm, circular permutations can be a bit tricky because rotations are considered identical. Let me recall how circular permutations work.

For a circular table, the number of ways to arrange n people is (n-1)! instead of n! because rotations are considered the same. So, if there were no restrictions, the answer would be (8-1)! = 7! = 5040 ways. But here, there's a restriction: Pierre, Rosa, and Thomas must all sit together.

So, how do I handle that? I think when certain people need to be together, we can treat them as a single unit or "block." Then, we can arrange the blocks around the table and then consider the arrangements within the block. But since it's a circular table, I need to adjust for that.

Let me think. If Pierre, Rosa, and Thomas are sitting together, we can think of them as one combined entity. So instead of 8 individuals, we have this block plus the other 5 people. That makes 6 entities in total. But wait, in circular permutations, arranging n entities around a table is (n-1)! So, if we have 6 entities, the number of ways to arrange them would be (6-1)! = 5! = 120. But then, within the block of Pierre, Rosa, and Thomas, they can be arranged among themselves in different ways. How many ways can they be arranged? Since there are 3 people, it's 3! = 6.

Therefore, the total number of arrangements would be 5! * 3! = 120 * 6 = 720. But wait, is that all? Let me check if there's something I'm missing here.

Wait, hold on. When dealing with circular permutations and blocks, sometimes there's a different way to account for rotations. Let me verify.

Suppose we fix the position of the block to eliminate rotational symmetry. For example, in circular permutations, fixing one position is a common technique. If we fix the block of Pierre, Rosa, and Thomas in a specific position, then we can arrange the remaining people relative to this block. But since rotations are considered the same, fixing the block's position would account for that.

Alternatively, if we consider the block as a single entity, then arranging the 6 entities around the table would normally be (6-1)! = 5! ways. Then, multiplying by the internal arrangements of the block, which is 3!. So, 5! * 3! = 720.

But let me think again. Suppose we have n objects around a circular table. The formula is (n-1)! because rotations are considered the same. So, if we have 6 objects (the block and the other 5 people), then the number of arrangements is (6-1)! = 120. Then, multiply by the number of ways to arrange the people inside the block, which is 3! = 6. So, 120 * 6 = 720.

But wait, another way to think about it: sometimes when you have a block, you can also consider linear arrangements. For example, if it were a straight line, the number of ways would be 6! * 3! because you have 6 units (the block and 5 others) arranged in a line, which is 6! ways, and then multiplied by 3! for the internal arrangements. But since it's a circular table, we need to adjust for rotational symmetry.

But in circular arrangements, fixing one position is equivalent to dividing by the number of rotations. However, when we have a block, does that affect the count? Let me check with a simpler case. Suppose there are 3 people who must sit together, and total of 4 people. How many circular arrangements?

Using the same method: treat the 3 people as a block, so we have 2 entities: the block and the 4th person. Number of circular arrangements is (2-1)! = 1! = 1. Multiply by internal arrangements: 3! = 6. So total 6.

But let's count manually. Let's say the people are A, B, C, D, and A, B, C must sit together. In circular table, fix D's position to eliminate rotations. Then, the block A-B-C can be arranged in 2 ways: clockwise or counter-clockwise? Wait, no. If D is fixed, then the block can be arranged in 3! ways, and the block can be placed either to the left or right of D? Wait, no. If D is fixed, then the block can be arranged in 3! ways, and since it's a circular table, once D is fixed, the block can be placed in only one position relative to D. Wait, but if we fix D's position, then the block can be arranged in 3! ways, and the rest is fixed. So total arrangements would be 3! = 6. Which matches the previous result. So that seems correct.

Alternatively, if we didn't fix D's position, how would it work? If we consider the block as a single entity, then we have 2 entities: block and D. The number of circular arrangements is (2-1)! = 1. Multiply by 3! for the block, so 6. So that works.

Therefore, going back to the original problem, treating the 3 people as a block, we have 6 entities, so (6-1)! = 120, times 3! = 6, so 720.

But wait, let me check another way. Suppose we first consider the total number of circular permutations without restrictions: 7! = 5040. Then, how many of these have Pierre, Rosa, and Thomas sitting together?

Alternatively, maybe using the formula for the number of circular permutations with a specific group together.

But I think the method I used earlier is correct. However, let me think again.

Another approach: In circular permutations, the number of ways where a specific group of k people are seated together is (n - k)! * k!

Wait, but in linear permutations, it's (n - k + 1)! * k!

Wait, no. For linear permutations, if you have n people and a group of k people must sit together, you treat the group as a single entity, so you have (n - k + 1) entities, which can be arranged in (n - k + 1)! ways, and multiplied by k! for the internal arrangements.

But for circular permutations, it's different.

So, for circular permutations, if we have a group of k people that must sit together, we can think of them as a single entity. Then, the number of entities is (n - k + 1). But since it's circular, the number of arrangements is (number of entities - 1)! So, (n - k + 1 - 1)! = (n - k)!

Then multiply by k! for the internal arrangements. Therefore, total number is (n - k)! * k!

Wait, so in our case, n = 8, k = 3. Therefore, (8 - 3)! * 3! = 5! * 3! = 120 * 6 = 720. Which matches the previous result.

Therefore, that seems to confirm that the answer is 720.

But let me check with another approach. Let's fix the position of the block to eliminate rotational symmetry.

Suppose we fix the block of Pierre, Rosa, and Thomas in a specific position. Since it's a circular table, we can fix their positions to account for rotations.

If we fix the block's position, then we can arrange the remaining 5 people in the remaining 5 seats. However, the block itself can be arranged in 3! ways.

But how many seats are there? The total number of seats is 8. If we fix the block of 3 people, then there are 8 - 3 = 5 seats left. But since we fixed the block's position, the rest of the people can be arranged in 5! ways.

Therefore, total number of arrangements is 3! * 5! = 6 * 120 = 720.

Same answer.

Alternatively, if we didn't fix the block's position, how would that work?

In circular permutations, fixing a position is a way to account for rotational symmetry. If we don't fix the block's position, then we have to consider that rotations are equivalent.

But when we consider the block as a single entity, the number of circular arrangements is (number of entities - 1)! So, with 6 entities, (6 - 1)! = 120. Multiply by 3! for the block's internal arrangements, 120 * 6 = 720.

So, all methods lead to 720.

But let me think again if there's a possibility of overcounting or undercounting.

Suppose we have the block of Pierre, Rosa, and Thomas. If we consider them as a single entity, then the total number of entities is 6. But in circular permutations, arranging 6 entities is (6 - 1)! = 120.

But another way: suppose we have 8 seats. Let's imagine that we first seat the block. Since it's a circular table, we can fix the block's position to remove rotational symmetry. So, fix Pierre, Rosa, and Thomas in seats 1, 2, 3. Then, the remaining 5 people can be arranged in seats 4 to 8, which is 5! ways. But within the block, Pierre, Rosa, and Thomas can be arranged in 3! ways. So total is 5! * 3! = 720.

Alternatively, if we don't fix the block's position, how would that work?

In circular permutations, the number of ways to arrange the block and the others is (number of entities - 1)! So, 5! as before.

But if we think of the block as a single entity, then the number of ways to arrange the block and the other 5 people is (6 - 1)! = 120. Then, multiply by 3! for the internal arrangements.

So, same answer.

Alternatively, if I think of the total number of ways where Pierre, Rosa, and Thomas are seated together as follows:

First, consider the number of ways to seat all 8 people around the table with Pierre, Rosa, and Thomas together.

First, let's think of the number of ways to seat them in a line, then adjust for circularity.

In linear arrangements, the number of ways where Pierre, Rosa, and Thomas are together is 6! * 3! as mentioned before. But for circular arrangements, it's different.

But how does that translate?

In linear arrangements, the number is (n - k + 1)! * k!

But for circular arrangements, it's (n - k)! * k!

Wait, so in our case, n = 8, k = 3, so (8 - 3)! * 3! = 5! * 3! = 720.

Alternatively, if I think of the circular table as a line with the ends connected. So, if we fix one person's position to make it linear.

For example, fix Pierre's position to eliminate rotational symmetry. Then, we need to arrange the remaining 7 people. But with the condition that Rosa and Thomas are sitting next to Pierre? Wait, no. Wait, the condition is that Pierre, Rosa, and Thomas are all sitting together.

So, if we fix Pierre's position, then Rosa and Thomas must be seated in the two seats adjacent to Pierre? Or they can be seated anywhere as long as all three are together.

Wait, if we fix Pierre's position, then the block of Pierre, Rosa, and Thomas can be arranged in two different ways: either Rosa and Thomas are on his left and right, or they are in some order. Wait, but since it's a circular table, fixing Pierre's position, the block can be arranged in different ways.

Wait, maybe this approach is more complicated. Let me try.

Fix Pierre's seat. Now, we need to arrange the other 7 people, with the condition that Rosa and Thomas are adjacent to Pierre or somewhere else? Wait, no. The condition is that all three are sitting together. So, Pierre, Rosa, and Thomas must form a block of three consecutive seats.

Since Pierre is fixed, the block can be either to his left or to his right. Wait, but since it's a circular table, fixing Pierre's position, the block can be in two different orientations?

Wait, no. If we fix Pierre's position, then the block of three people can be arranged in two different ways: either the block is Pierre, Rosa, Thomas in that order, or Pierre, Thomas, Rosa, or other permutations. Wait, but the block can be placed in different positions.

Wait, if Pierre is fixed, then the block of three can be seated in two different positions: either the block extends clockwise from Pierre, or counter-clockwise from Pierre. But since the block is three people, if Pierre is fixed, then the block can be in positions (Pierre, seat 2, seat 3) or (seat 8, seat 1, Pierre). Wait, but seat 1 is Pierre's seat. So, if we fix Pierre at seat 1, then the block can be seats 1, 2, 3 or seats 8, 1, 2. But seat 1 is already occupied by Pierre. So, the block can be either seats 1, 2, 3 or seats 8, 1, 2. However, seat 1 is Pierre, so in the first case, the block is Pierre, seat 2, seat 3. In the second case, seat 8, Pierre, seat 2.

But in both cases, the block includes Pierre. So, the block can be in two different positions relative to Pierre's fixed seat. However, since we fixed Pierre's seat, the rest of the block can be arranged in different ways.

But maybe this is getting too complicated. Let me think.

If we fix Pierre's seat, then the number of ways to arrange the rest with Rosa and Thomas adjacent to Pierre? No, not necessarily adjacent. The block can be anywhere as long as all three are together.

Wait, if Pierre is fixed, then the block of three can be placed in different positions. For example, the block could be Pierre, Rosa, Thomas in seats 1, 2, 3; or seats 1, 8, 7; or seats 1, 2, 8? Wait, no. Wait, the block has to be three consecutive seats.

Since the table is circular, the number of ways to place a block of three consecutive seats with Pierre fixed at seat 1.

If Pierre is at seat 1, then the block can start at seat 1, 2, or 8? Wait, no. If the block is three consecutive seats, then starting at seat 1, the block would be seats 1, 2, 3. Starting at seat 8, the block would be seats 8, 1, 2. Starting at seat 7, the block would be seats 7, 8, 1. But seat 1 is already occupied by Pierre. So, if we want the block to include Pierre, then the block can be seats 8, 1, 2 or seats 1, 2, 3. So, two possible blocks that include Pierre.

But if we don't require the block to include Pierre? Wait, no. The block must include Pierre, Rosa, and Thomas. Since Pierre is fixed, the block must include Pierre. Therefore, the block can be in two different positions: either seats 1, 2, 3 or seats 8, 1, 2.

But in each case, the block includes Pierre. So, for each of these two positions, we can arrange Rosa and Thomas in the other two seats of the block.

So, for each of the two positions (block starting at seat 1 or block starting at seat 8), we can arrange Rosa and Thomas in 2! ways.

But also, the block can be arranged in different orders. Wait, if the block is seats 1, 2, 3, then Pierre is in seat 1, and Rosa and Thomas can be in seats 2 and 3 in 2! ways. Similarly, if the block is seats 8, 1, 2, then Pierre is in seat 1, and Rosa and Thomas can be in seats 8 and 2 in 2! ways.

But also, the block could be arranged in different orders. Wait, for example, if the block is seats 1, 2, 3, Pierre could be in seat 1, 2, or 3? Wait, no, because we fixed Pierre at seat 1. Therefore, Pierre must be in seat 1, and the other two seats in the block (2 and 3) are for Rosa and Thomas. Similarly, if the block is seats 8, 1, 2, then Pierre is in seat 1, and the other two seats (8 and 2) are for Rosa and Thomas.

But wait, if we fix Pierre at seat 1, then the block can be in two different positions (either extending clockwise or counter-clockwise from Pierre). However, in each case, the other two seats in the block are adjacent to Pierre. Therefore, the number of ways to arrange Rosa and Thomas in the block is 2! for each block position.

But there are two block positions, so total arrangements for the block would be 2 * 2! = 4.

Then, after placing the block, we have 5 remaining people to arrange in the remaining 5 seats. Since we fixed Pierre's position, the rest can be arranged in 5! ways.

Therefore, total number of arrangements would be 4 * 5! = 4 * 120 = 480.

Wait, but this contradicts the previous answer of 720. What's going on here?

Hmm, so according to this method, fixing Pierre's position gives 480, but earlier methods gave 720. There's a discrepancy here. Which one is correct?

Let me check.

If I fix Pierre's position, then the number of ways where Pierre, Rosa, and Thomas are all together is 2 * 2! * 5! = 480. But according to the previous methods, it's 720.

So, which is correct?

Alternatively, maybe I made a mistake in this approach. Let me think again.

If we fix Pierre's position, then the block of three can be placed in different ways. However, maybe there are more positions for the block?

Wait, if the block is three consecutive seats, and Pierre is fixed at seat 1, then the block can be:

- Seats 1, 2, 3

- Seats 8, 1, 2

- Seats 7, 8, 1

Wait, that's three different blocks that include seat 1. Wait, seat 7, 8, 1 is another block that includes seat 1. So, actually, there are three different blocks that include seat 1.

Wait, let me visualize the table. Let's label the seats 1 through 8 clockwise. If Pierre is fixed at seat 1, then the blocks of three consecutive seats that include seat 1 are:

1. Seats 1, 2, 3

2. Seats 8, 1, 2

3. Seats 7, 8, 1

So, three different blocks. Therefore, there are three positions for the block that includes seat 1.

Therefore, for each of these three blocks, we can arrange Pierre, Rosa, and Thomas in the block. However, since Pierre is fixed at seat 1, the block must include seat 1.

So, for each of these three blocks, we need to arrange Rosa and Thomas in the other two seats of the block.

For example:

- Block 1, 2, 3: Pierre is at 1, so Rosa and Thomas can be at 2 and 3 in 2! ways.

- Block 8, 1, 2: Pierre is at 1, so Rosa and Thomas can be at 8 and 2 in 2! ways.

- Block 7, 8, 1: Pierre is at 1, so Rosa and Thomas can be at 7 and 8 in 2! ways.

Therefore, total number of ways to arrange the block is 3 * 2! = 6.

Then, the remaining 5 people can be arranged in the remaining 5 seats in 5! ways.

Therefore, total number of arrangements is 6 * 5! = 6 * 120 = 720.

Ah! So earlier, I had only considered two blocks, but there are actually three blocks that include seat 1. Therefore, the correct number is 3 * 2! * 5! = 720, which matches the previous answer.

So, my mistake earlier was not considering all possible blocks that include Pierre's fixed seat.

Therefore, this method also gives 720.

Therefore, this confirms that the answer is 720.

But let me check once more with another approach.

Suppose we don't fix any positions.

The total number of circular permutations where Pierre, Rosa, and Thomas are seated together is equal to the number of ways to arrange the 8 people around the table with those three together.

As mentioned before, treating them as a single block, we have 6 entities. The number of circular arrangements is (6 - 1)! = 120. Multiply by the number of ways to arrange the block, which is 3! = 6. So, 120 * 6 = 720.

Alternatively, using the formula for circular permutations with a group together: (n - k)! * k! where n is total people, k is the size of the group.

So, (8 - 3)! * 3! = 5! * 3! = 120 * 6 = 720.

Therefore, all methods lead to 720.

But let me think if there's another way someone might approach this problem and get a different answer.

For example, someone might think: first, choose the seats for Pierre, Rosa, and Thomas. Since they must sit together, they occupy 3 consecutive seats. How many ways are there to choose 3 consecutive seats around a circular table?

In a circular table with 8 seats, the number of ways to choose a block of 3 consecutive seats is 8. Because for each seat, there is a block starting at that seat. However, since the table is circular, each block is counted once. Wait, for example, seats 1-2-3, 2-3-4,..., 8-1-2. So, there are 8 different blocks of 3 consecutive seats.

But if we consider that each block is unique, then once we choose a block, we can arrange Pierre, Rosa, and Thomas in 3! ways. Then, arrange the remaining 5 people in the remaining 5 seats in 5! ways.

But since the table is circular, are we overcounting?

Wait, if we consider that choosing different blocks might lead to equivalent arrangements due to rotation.

But if we fix the block, then we have to consider that rotations are considered the same.

Wait, if we first count the number of ways as follows:

Number of ways = (number of blocks) * (arrangements within block) * (arrangements of others).

But since rotations are considered the same, how does that affect?

If we count the number of blocks as 8, then multiply by 3! * 5!, we would get 8 * 6 * 120 = 5760. But this is way larger than the total number of circular permutations, which is 7! = 5040.

Therefore, clearly, this approach is wrong.

So, why is that? Because when we count the number of blocks as 8, we are considering different rotations as different blocks, but in reality, in circular permutations, rotations are considered the same. Therefore, we need to adjust for that.

Therefore, if we use this method, we have to divide by the number of rotations.

Wait, but how?

Alternatively, if we consider that each unique circular arrangement is counted multiple times in the block count.

For example, each unique arrangement where the block is seated somewhere is counted once for each rotation.

But since the block can be rotated, but the entire arrangement is considered the same.

This seems complicated.

Alternatively, since we know that the total number of circular arrangements is 7! = 5040, and the number of arrangements where the block is seated together is 720, then 720 / 5040 = 1/7. Which makes sense, because there are 8 possible positions for the block, but since rotations are considered the same, each unique arrangement is counted 8 times? Wait, no.

Wait, if we fix the block, then the number of unique arrangements is (number of blocks) * ...

But this seems messy.

Alternatively, since we know that the answer is 720, and the total number of circular permutations is 5040, then the probability that a random seating has the three people together is 720 / 5040 = 1/7.

But is that correct?

Alternatively, the number of ways to seat the three people together is (number of ways to seat them as a block) = (n - 1)! * k! / n? Wait, not sure.

But since we have already confirmed with multiple methods that the answer is 720, and the alternative approach of counting blocks led to overcounting, I think the correct answer is 720.

Therefore, after all these verifications, I think the answer is 720.

But just to make sure, let me think of a smaller case. Let's take n=3, and the group size k=3. Then, the number of circular arrangements where all three are together is 1 (since they are all together). According to the formula (n - k)! * k! = (3 - 3)! * 3! = 1 * 6 = 6. But that's wrong because in circular permutations, there's only 1 way to arrange 3 people.

Wait, this is a problem.

Wait, hold on. If n=3, and k=3, then according to the formula (n - k)! * k! = 0! * 3! = 1 * 6 = 6. But the actual number of circular permutations is (3 - 1)! = 2. So, clearly, the formula is wrong here.

Wait, so my previous reasoning must have a mistake.

Wait, so what's the problem here?

In the case of n=3, k=3, the number of circular arrangements where all three are together is 1 (since they are all together). But according to the formula (n - k)! * k! = 1 * 6 = 6. Which is wrong.

So, there's a mistake in my previous reasoning.

Wait, so where is the error?

Earlier, I thought that the formula is (n - k)! * k! for circular permutations with a group of k people together. But in this case, it's not.

So, let's think again.

For n=3, k=3. How many circular arrangements where all three are together? Well, since they are all together, it's just the number of circular permutations of 3 people, which is (3 - 1)! = 2. But according to the formula (n - k)! * k! = 1 * 6 = 6. Which is wrong.

Therefore, my previous formula is incorrect.

So, what's the correct formula?

Let me think again.

When dealing with circular permutations where a group of k people must sit together, the correct approach is to treat the group as a single entity. Then, the number of entities is (n - k + 1). Wait, no.

Wait, if we have n people, and we treat a group of k as a single entity, then the number of entities is (n - k + 1)? No, it's (n - k + 1) if we are considering linear arrangements.

Wait, no. If you have n people and you group k of them together, you have (n - k + 1) entities? No, that's for linear arrangements.

Wait, in linear arrangements, the number of ways to arrange n people with a group of k together is (n - k + 1)! * k!

But for circular arrangements, it's different.

Let me think again.

In circular arrangements, when we have a group of k people that must sit together, we can consider them as a single entity. Then, the number of entities is (n - k + 1)? No, it's (n - k + 1) if we are considering linear arrangements.

Wait, no. If we have n people, and we group k of them into a single entity, then the number of entities is (n - k + 1)? No, it's (n - k + 1) only if we are considering the group as a single entity and the rest as individuals. Wait, no.

Actually, if you have n people and you group k of them into a single entity, then the number of entities is (n - k) + 1 = n - k + 1.

But in circular permutations, the number of ways to arrange these entities is (number of entities - 1)!

Therefore, total number of arrangements is (n - k + 1 - 1)! * k! = (n - k)! * k!

But in the case of n=3, k=3, this gives (3 - 3)! * 3! = 1 * 6 = 6, which is wrong.

But the correct number is 2.

So, clearly, this formula is not correct.

Therefore, my previous reasoning was flawed.

So, what is the correct formula?

Let me think again.

In the case of n=3, k=3. If we treat the group as a single entity, then we have 1 entity. The number of circular arrangements of 1 entity is (1 - 1)! = 0! = 1. Multiply by the internal arrangements of the group, which is 3! = 6. So total is 1 * 6 = 6. But the actual number of circular arrangements is 2.

So, there's a contradiction here.

Wait, why is that?

Because when we treat the group as a single entity, we are assuming that the group can be placed anywhere, but in reality, when the group is the entire set, the circular arrangement is just the number of circular permutations of the entire group.

But in this case, the formula gives 6, but the correct answer is 2.

So, the problem is that when the group is the entire set, the formula overcounts.

Wait, but why?

Because when we treat the group as a single entity, we are considering that the group can be arranged in k! ways, but in reality, in circular permutations, the entire group's arrangement is already considered.

Wait, maybe the formula is correct, but the interpretation is wrong.

Wait, if we have n=3, and we treat the group of 3 as a single entity, then the number of circular arrangements is (1 - 1)! * 3! = 1 * 6 = 6. But in reality, the number of circular arrangements of 3 people is 2.

So, why is there a discrepancy?

Because when we treat the group as a single entity, we are not accounting for the fact that the entire group is the whole table.

Wait, in other words, when we have a single entity, the number of circular arrangements is 1 (since there's only one entity), but we multiply by the internal arrangements. However, in reality, the internal arrangements are already considered in the circular permutations.

Wait, no. In circular permutations, when you have a single entity, it's just one way to arrange it, but the internal arrangements are separate.

But in reality, arranging the single entity (which is the entire group) around the table is just one way, but the internal arrangements are the permutations of the group. However, in circular permutations, the internal arrangements of the group would be considered as linear arrangements, but since the entire table is circular, does that affect it?

Wait, for example, if you have three people around a table, the number of circular permutations is 2. However, if you consider them as a single entity, then the number of circular arrangements is 1 (for the entity), multiplied by 3! (internal arrangements). But 1 * 6 = 6, which is not equal to 2.

Therefore, there's a problem here.

So, what is the correct way to think about this?

Perhaps the formula (n - k)! * k! is not correct for circular permutations.

Let me think again.

In the case of n=3, k=3, the correct answer is 2.

If we use the method of fixing the block's position.

Fix one person's position to eliminate rotational symmetry. Let's fix Pierre at seat 1. Then, the other two people, Rosa and Thomas, can be arranged in 2! ways. So total is 2. Which matches the correct answer.

But according to the formula (n - k)! * k! = 1 * 6 = 6, which is wrong.

Therefore, the formula is incorrect.

So, what is the correct formula?

Let me think again.

When we have a group of k people that must sit together, we can treat them as a single entity. Then, the number of entities is (n - k + 1)? No, it's (n - k + 1) if we are considering linear arrangements.

Wait, no. If we have n people and we group k of them into a single entity, then the number of entities is (n - k) + 1 = n - k + 1.

But in circular arrangements, the number of ways to arrange these entities is (number of entities - 1)!

Therefore, total number of arrangements is (n - k + 1 - 1)! * k! = (n - k)! * k!

But this gives the wrong answer for n=3, k=3.

So, why?

Because when the group is the entire set, the formula counts the internal arrangements as k! but in reality, the circular permutations already account for rotational symmetry.

Wait, maybe the formula is correct, but when we fix the group as a single entity, we need to adjust for rotational symmetry.

Wait, if we have n=3, k=3, then the number of entities is 1. The number of circular arrangements of 1 entity is 1 (since there's only one way to arrange one entity around a table). Multiply by the internal arrangements of the entity, which is 3! = 6. But in reality, the number of circular arrangements is 2.

So, there's a contradiction.

Wait, but if we think of the internal arrangements as linear arrangements, then maybe we need to divide by something?

Wait, no. In circular permutations, when you have a single entity, the internal arrangements are linear?

Wait, no. If the entire group is the single entity, then the internal arrangements are the circular permutations of the group?

Wait, no. If the entire group is considered as a single entity, then the internal arrangements are the permutations of the group, but since the entire table is circular, the internal arrangements are actually linear?

This is confusing.

Let me think of another example. Suppose n=4, k=2. How many circular arrangements where two specific people are together?

Using the formula (n - k)! * k! = (4 - 2)! * 2! = 2! * 2! = 4.

But let's count manually.

Total circular permutations of 4 people: (4 - 1)! = 6.

Number of circular arrangements where two specific people are together:

Treat the two people as a block. Then, we have 3 entities: the block and the other two people. Number of circular arrangements: (3 - 1)! = 2. Multiply by internal arrangements of the block: 2! = 2. Total: 2 * 2 = 4. Which matches the formula.

But let's count manually.

Fix one person, say Pierre, to eliminate rotational symmetry. Then, the other person (Rosa) can be either to his left or right. So, 2 choices. Then, the remaining two people can be arranged in 2! ways. So total is 2 * 2 = 4. Which matches.

So, in this case, the formula works.

But in the case of n=3, k=3, the formula gives 6, but the correct answer is 2.

Wait, why is that?

If we use the same method: fix one person, say Pierre. Then, the other two people (Rosa and Thomas) can be arranged in 2! ways. So total is 2. Which is correct.

But according to the formula (n - k)! * k! = 1 * 6 = 6.

So, why is there a discrepancy?

Ah! Because when we treat the group as a single entity, and calculate (number of entities - 1)! * k! = (1 - 1)! * 3! = 1 * 6 = 6, but in reality, when we fix one person, we are effectively accounting for rotational symmetry, and the formula should be adjusted.

Wait, so maybe the formula is not (n - k)! * k! but something else.

Wait, in the case of n=3, k=3, if we treat the group as a single entity, then the number of circular arrangements is (number of entities - 1)! = 0! = 1, and then multiply by the internal arrangements of the group, which is 3! = 6. But in reality, the number of circular arrangements is 2.

So, why is there a difference?

Because when we treat the group as a single entity, we are not accounting for the fact that the entire group is the whole table.

Wait, if we have a single entity, which is the entire group, then arranging it around the table is just 1 way, but the internal arrangements are 3! = 6. However, in reality, the circular permutations of the entire group is (3 - 1)! = 2.

So, the formula is overcounting by a factor of 3.

Wait, why?

Because when we treat the group as a single entity, we are considering the internal arrangements as linear, but in reality, they should be circular.

Ah! So, the mistake is that when we treat the group as a single entity, we should consider the internal arrangements as circular permutations, not linear.

Wait, but the group is a block that is part of the circular table.

Wait, this is getting too tangled.

Let me think again.

If we have a group of k people that must sit together, and we treat them as a single entity, then the number of ways to arrange the entities around the table is (number of entities - 1)!

But the internal arrangements of the group: if the group is a block, then their internal arrangements are linear, because the block is part of the circular table.

Wait, no. If the block is part of the circular table, then the internal arrangements of the block are linear? Or are they circular?

Actually, the internal arrangements of the block are linear because the block is a consecutive sequence. For example, if the block is three people sitting together, their arrangement is linear relative to the rest of the table.

Wait, but in reality, the entire table is circular, so the block's internal arrangement is a linear arrangement, because the block is a consecutive segment.

Therefore, the internal arrangements should be k!

But in the case of n=3, k=3, the internal arrangements would be 3! = 6, but the actual number of circular permutations is 2.

So, why is there a discrepancy?

Because when the entire group is the table, the internal arrangements are actually circular.

Wait, if the entire group is the table, then the internal arrangements are circular. So, in that case, the number of internal arrangements should be (k - 1)! instead of k!

Ah! So, the formula should be:

If we have a group of k people that must sit together, treat them as a single entity. The number of circular arrangements is (number of entities - 1)! * (k - 1)!

Wait, let's test this.

For n=3, k=3: number of entities = 1. So, (1 - 1)! * (3 - 1)! = 1 * 2 = 2, which is correct.

For n=4, k=2: number of entities = 2 (the block and the other two people). Wait, no. If n=4, k=2, then number of entities is (4 - 2) + 1 = 3? Wait, no.

Wait, if we have n=4 people, and we group k=2 into a block, then the number of entities is 3: the block and the other two individuals.

So, number of circular arrangements is (3 - 1)! * (2 - 1)! = 2! * 1! = 2 * 1 = 2. But earlier, we saw that the correct number is 4.

Wait, this is conflicting.

Wait, earlier, for n=4, k=2, the correct number of circular arrangements where the two people are together is 4. But according to this new formula, it would be (number of entities - 1)! * (k - 1)! = (3 - 1)! * (2 - 1)! = 2! * 1! = 2. Which is wrong.

So, this suggests that this formula is also incorrect.

This is getting too confusing. Let me try to find a general formula.

Let me refer back to standard combinatorial results.

In circular permutations, the number of ways to arrange n people with a specific group of k people together is (n - k)! * k!

Wait, but this gave the wrong answer for n=3, k=3.

But in the case of n=4, k=2, it gave 2! * 2! = 4, which is correct.

Wait, for n=3, k=3, it gives 0! * 3! = 1 * 6 = 6, which is wrong.

But why is it correct for n=4, k=2 but not for n=3, k=3?

Maybe because when the group is the entire set, the formula overcounts.

Wait, but in the case of n=3, k=3, the formula gives 6, but the correct answer is 2.

But if we think of the formula as (n - k)! * k! = (3 - 3)! * 3! = 1 * 6 = 6, but the actual number of circular permutations is 2.

So, the formula is overcounting by a factor of 3.

Wait, but why?

Because when we treat the group as a single entity, we are considering the internal arrangements as linear, but in reality, when the group is the entire set, the internal arrangements should be circular.

Wait, but in the case of n=4, k=2, the group is not the entire set, so the internal arrangements are linear.

So, perhaps the formula is correct when the group is not the entire set, but when the group is the entire set, we need to adjust.

But this seems inconsistent.

Alternatively, maybe the formula is correct, and the problem is in the interpretation of the question.

Wait, in the original problem, the question says: "Two seatings are considered the same if one is a rotation of the other."

So, in the case of n=3, k=3, the number of distinct seatings is 2, but according to the formula (n - k)! * k! = 6, which suggests that there are 6 distinct seatings, which is wrong.

But why?

Because when we use the formula (n - k)! * k!, we are assuming that the group is treated as a single entity, and the rest are individuals, and then arranging them around the table.

But in the case where the group is the entire set, this would mean arranging 1 entity (the group) around the table, which is 1 way, multiplied by k! = 6, giving 6. But in reality, the number of distinct seatings is 2.

So, the formula is wrong.

But why does it work for n=4, k=2?

Let me think. For n=4, k=2:

Using the formula: (4 - 2)! * 2! = 2! * 2! = 4.

Which matches the correct answer.

But if we use the same logic as for n=3, k=3, why is it different?

Maybe because when the group is not the entire set, the formula works, but when it is, it doesn't.

This suggests that the formula is not universally applicable.

Alternatively, perhaps the formula is correct, but the interpretation of the problem is different.

Wait, in the case of n=3, k=3, if we consider the formula (n - k)! * k! = 6, but the actual number of distinct seatings is 2, this suggests that the formula counts each seating multiple times.

Wait, how many times is each seating counted in the formula?

If we consider that each seating is counted k times, because the group can be rotated.

Wait, for n=3, k=3, each seating is counted 3 times in the formula.

Because the formula counts all possible rotations of the group.

But since rotations are considered the same, we need to divide by the number of rotations.

Wait, but in the formula, we already accounted for rotational symmetry by using (number of entities - 1)!

Wait, this is getting too tangled.

Maybe I should refer to standard references.

Wait, according to standard combinatorics, the number of ways to arrange n people around a circular table with a specific group of k people together is (n - k)! * k!

But this seems to conflict with the n=3, k=3 case.

Wait, let me check an external source.

Wait, I can't access external sources, but I can think.

In the case of circular permutations, when we have a group of k people that must sit together, the standard approach is to treat the group as a single entity, leading to (n - k + 1 - 1)! * k! = (n - k)! * k!

Wait, but this is the same as before.

But in the case of n=3, k=3, this gives 1 * 6 = 6, but the correct answer is 2.

So, what is the issue?

Ah! I think the issue is that when the group is the entire set, the formula counts the number of linear arrangements, not circular.

Wait, no.

Wait, let's think of it this way: when you treat the group as a single entity, you are effectively converting the circular permutation problem into a linear one.

For example, if you have a circular table with n people, and you fix one person's position to make it linear, then the number of arrangements is (n - 1)!

But if you have a group of k people, you can fix the group's position to make it linear.

Wait, this is getting too abstract.

Let me try to think of the n=3, k=3 case.

If we have 3 people around a table, and we want to count the number of circular arrangements where all three are together. Well, they are always together, so it's just the number of circular permutations, which is 2.

But according to the formula (n - k)! * k! = 1 * 6 = 6.

So, why is there a discrepancy?

Because when we treat the group as a single entity, we are not accounting for the fact that the group's internal arrangements are also circular.

Wait, if the group is a single entity, then its internal arrangements are circular?

No, the group is a block of people sitting together, so their internal arrangement is linear.

Wait, but in the case of the entire group, the block is the whole table, so their internal arrangement is circular.

Therefore, in this case, the internal arrangements should be (k - 1)! instead of k!

So, if we adjust the formula to (n - k)! * (k - 1)!

For n=3, k=3: (3 - 3)! * (3 - 1)! = 1 * 2 = 2, which is correct.

For n=4, k=2: (4 - 2)! * (2 - 1)! = 2! * 1! = 2 * 1 = 2, which is wrong, because earlier we saw that the correct number is 4.

Wait, now it's conflicting again.

This suggests that the formula is not consistent.

This is very confusing.

Let me try to think of the general case.

When we have a group of k people that must sit together, we can treat them as a single entity. The number of ways to arrange the entities around the table is (number of entities - 1)!

The number of ways to arrange the group internally is k! if the group is treated as a linear block, or (k - 1)! if the group is treated as a circular block.

But in the context of the entire table being circular, the group's internal arrangement is linear, because the group is part of the circular table.

Wait, for example, if you have a block of k people sitting together, their arrangement is linear relative to the rest of the table.

Therefore, the internal arrangements should be k!

But in the case where the entire table is the group, the internal arrangement is circular, so it should be (k - 1)!

This suggests that the formula depends on whether the group is the entire table or not.

This seems too complicated.

Perhaps the correct approach is to use the method of fixing a position to eliminate rotational symmetry.

For example, in the original problem, with n=8, k=3.

Fix one person's position, say Pierre. Then, the number of ways to arrange the rest with Rosa and Thomas sitting together.

Wait, but this is what I did earlier, and it gave 720.

But in the case of n=3, k=3, fixing Pierre's position, the number of ways is 2, which is correct.

So, perhaps the correct formula is:

Number of circular arrangements = (number of ways to arrange the rest with the group together) = (n - 1)! * (k / n) ?

No, not sure.

Alternatively, the number of ways is (n - 1)! * (k / n) ?

Wait, for n=3, k=3: (3 - 1)! * (3 / 3) = 2 * 1 = 2, which is correct.

For n=4, k=2: (4 - 1)! * (2 / 4) = 6 * 0.5 = 3, which is wrong, because the correct number is 4.

Hmm, no.

Alternatively, the number of ways is (n - 1)! * (k! / n) ?

For n=3, k=3: 2! * (6 / 3) = 2 * 2 = 4, which is wrong.

This is not working.

Let me go back to the original problem.

Since in the original problem, using the method of fixing Pierre's position gave 720, and other methods also gave 720, and in the n=3, k=3 case, fixing one person's position gives the correct answer, perhaps the formula is:

Number of circular arrangements = (n - 1)! * (k! / n) ?

Wait, no.

Wait, in the original problem, if we fix Pierre's position, then the number of ways is:

Number of ways to arrange the rest with Rosa and Thomas together.

Which, as we calculated earlier, is 3 * 2! * 5! = 720.

But (n - 1)! is 7! = 5040.

So, 720 is 5040 * (720 / 5040) = 5040 * (1/7).

So, the probability is 1/7.

But why 1/7?

Because there are 8 possible positions for the block of 3 people, but since rotations are considered the same, each unique arrangement is counted 8 times?

Wait, no.

Alternatively, the number of ways to choose the block is 8, but each unique arrangement is counted 3 times (since the block can start at any of the 3 positions).

Wait, this is getting too tangled.

But since in the original problem, all methods gave 720, and in the n=3, k=3 case, fixing one person's position gives the correct answer, I think that the correct answer for the original problem is 720.

Therefore, despite the confusion with the n=3, k=3 case, I think that the original answer of 720 is correct.

Perhaps the formula (n - k)! * k! is correct when the group is not the entire set, and in the case where the group is the entire set, it's a special case.

But since in the original problem, the group is not the entire set, and all methods give 720, I will go with 720 as the answer.

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

**Final Answer**
The number of ways is \boxed{720}.

To determine the number of ways 8 people can sit around a round table such that **Pierre, Rosa, and Thomas all sit together**, we need to carefully account for both the **circular nature of the table** and the **restriction that three specific people must be adjacent**.

---

### Step 1: Understanding Circular Permutations

In a **circular arrangement**, the number of distinct ways to arrange $ n $ people is:

$$
(n - 1)!
$$

This is because rotations of the same arrangement are considered identical. For example, the arrangement ABCDE is the same as BCDEA in a circular table.

---

### Step 2: Treating the Group as a Single Entity

Since **Pierre, Rosa, and Thomas** must sit together, we can treat them as a **single block** or **entity**. This reduces the problem to arranging:

- 1 block (Pierre, Rosa, Thomas)
- 5 other individuals

So, we now have **6 entities** to arrange around the table.

---

### Step 3: Arranging the 6 Entities

The number of ways to arrange 6 entities in a circle is:

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

---

### Step 4: Arranging the People Within the Block

Within the block of 3 people (Pierre, Rosa, and Thomas), they can be arranged among themselves in:

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

---

### Step 5: Total Number of Arrangements

Now, multiply the number of ways to arrange the 6 entities by the number of internal arrangements within the block:

$$
5! \times 3! = 120 \times 6 = 720
$$

---

### 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. And rotations are considered the same seating. Hmm, circular permutations can be a bit tricky because rotations are considered identical. Let me recall how circular permutations work.

For a circular table, the number of ways to arrange n people is (n-1)! instead of n! because rotations are considered the same. So, if there were no restrictions, the answer would be (8-1)! = 7! = 5040 ways. But here, there's a restriction: Pierre, Rosa, and Thomas must all sit together.

So, how do I handle that? I think when certain people need to be together, we can treat them as a single unit or "block." Then, we can arrange the blocks around the table and then consider the arrangements within the block. But since it's a circular table, I need to adjust for that.

Let me think. If Pierre, Rosa, and Thomas are sitting together, we can think of them as one combined entity. So instead of 8 individuals, we have this block plus the other 5 people. That makes 6 entities in total. But wait, in circular permutations, arranging n entities around a table is (n-1)! So, if we have 6 entities, the number of ways to arrange them would be (6-1)! = 5! = 120. But then, within the block of Pierre, Rosa, and Thomas, they can be arranged among themselves in different ways. How many ways can they be arranged? Since there are 3 people, it's 3! = 6.

Therefore, the total number of arrangements would be 5! * 3! = 120 * 6 = 720. But wait, is that all? Let me check if there's something I'm missing here.

Wait, hold on. When dealing with circular permutations and blocks, sometimes there's a different way to account for rotations. Let me verify.

Suppose we fix the position of the block to eliminate rotational symmetry. For example, in circular permutations, fixing one position is a common technique. If we fix the block of Pierre, Rosa, and Thomas in a specific position, then we can arrange the remaining people relative to this block. But since rotations are considered the same, fixing the block's position would account for that.

Alternatively, if we consider the block as a single entity, then arranging the 6 entities around the table would normally be (6-1)! = 5! ways. Then, multiplying by the internal arrangements of the block, which is 3!. So, 5! * 3! = 720.

But let me think again. Suppose we have n objects around a circular table. The formula is (n-1)! because rotations are considered the same. So, if we have 6 objects (the block and the other 5 people), then the number of arrangements is (6-1)! = 120. Then, multiply by the number of ways to arrange the people inside the block, which is 3! = 6. So, 120 * 6 = 720.

But wait, another way to think about it: sometimes when you have a block, you can also consider linear arrangements. For example, if it were a straight line, the number of ways would be 6! * 3! because you have 6 units (the block and 5 others) arranged in a line, which is 6! ways, and then multiplied by 3! for the internal arrangements. But since it's a circular table, we need to adjust for rotational symmetry.

But in circular arrangements, fixing one position is equivalent to dividing by the number of rotations. However, when we have a block, does that affect the count? Let me check with a simpler case. Suppose there are 3 people who must sit together, and total of 4 people. How many circular arrangements?

Using the same method: treat the 3 people as a block, so we have 2 entities: the block and the 4th person. Number of circular arrangements is (2-1)! = 1! = 1. Multiply by internal arrangements: 3! = 6. So total 6.

But let's count manually. Let's say the people are A, B, C, D, and A, B, C must sit together. In circular table, fix D's position to eliminate rotations. Then, the block A-B-C can be arranged in 2 ways: clockwise or counter-clockwise? Wait, no. If D is fixed, then the block can be arranged in 3! ways, and the block can be placed either to the left or right of D? Wait, no. If D is fixed, then the block can be arranged in 3! ways, and since it's a circular table, once D is fixed, the block can be placed in only one position relative to D. Wait, but if we fix D's position, then the block can be arranged in 3! ways, and the rest is fixed. So total arrangements would be 3! = 6. Which matches the previous result. So that seems correct.

Alternatively, if we didn't fix D's position, how would it work? If we consider the block as a single entity, then we have 2 entities: block and D. The number of circular arrangements is (2-1)! = 1. Multiply by 3! for the block, so 6. So that works.

Therefore, going back to the original problem, treating the 3 people as a block, we have 6 entities, so (6-1)! = 120, times 3! = 6, so 720.

But wait, let me check another way. Suppose we first consider the total number of circular permutations without restrictions: 7! = 5040. Then, how many of these have Pierre, Rosa, and Thomas sitting together?

Alternatively, maybe using the formula for the number of circular permutations with a specific group together.

But I think the method I used earlier is correct. However, let me think again.

Another approach: In circular permutations, the number of ways where a specific group of k people are seated together is (n - k)! * k!

Wait, but in linear permutations, it's (n - k + 1)! * k!

Wait, no. For linear permutations, if you have n people and a group of k people must sit together, you treat the group as a single entity, so you have (n - k + 1) entities, which can be arranged in (n - k + 1)! ways, and multiplied by k! for the internal arrangements.

But for circular permutations, it's different.

So, for circular permutations, if we have a group of k people that must sit together, we can think of them as a single entity. Then, the number of entities is (n - k + 1). But since it's circular, the number of arrangements is (number of entities - 1)! So, (n - k + 1 - 1)! = (n - k)!

Then multiply by k! for the internal arrangements. Therefore, total number is (n - k)! * k!

Wait, so in our case, n = 8, k = 3. Therefore, (8 - 3)! * 3! = 5! * 3! = 120 * 6 = 720. Which matches the previous result.

Therefore, that seems to confirm that the answer is 720.

But let me check with another approach. Let's fix the position of the block to eliminate rotational symmetry.

Suppose we fix the block of Pierre, Rosa, and Thomas in a specific position. Since it's a circular table, we can fix their positions to account for rotations.

If we fix the block's position, then we can arrange the remaining 5 people in the remaining 5 seats. However, the block itself can be arranged in 3! ways.

But how many seats are there? The total number of seats is 8. If we fix the block of 3 people, then there are 8 - 3 = 5 seats left. But since we fixed the block's position, the rest of the people can be arranged in 5! ways.

Therefore, total number of arrangements is 3! * 5! = 6 * 120 = 720.

Same answer.

Alternatively, if we didn't fix the block's position, how would that work?

In circular permutations, fixing a position is a way to account for rotational symmetry. If we don't fix the block's position, then we have to consider that rotations are equivalent.

But when we consider the block as a single entity, the number of circular arrangements is (number of entities - 1)! So, with 6 entities, (6 - 1)! = 120. Multiply by 3! for the block's internal arrangements, 120 * 6 = 720.

So, all methods lead to 720.

But let me think again if there's a possibility of overcounting or undercounting.

Suppose we have the block of Pierre, Rosa, and Thomas. If we consider them as a single entity, then the total number of entities is 6. But in circular permutations, arranging 6 entities is (6 - 1)! = 120.

But another way: suppose we have 8 seats. Let's imagine that we first seat the block. Since it's a circular table, we can fix the block's position to remove rotational symmetry. So, fix Pierre, Rosa, and Thomas in seats 1, 2, 3. Then, the remaining 5 people can be arranged in seats 4 to 8, which is 5! ways. But within the block, Pierre, Rosa, and Thomas can be arranged in 3! ways. So total is 5! * 3! = 720.

Alternatively, if we don't fix the block's position, how would that work?

In circular permutations, the number of ways to arrange the block and the others is (number of entities - 1)! So, 5! as before.

But if we think of the block as a single entity, then the number of ways to arrange the block and the other 5 people is (6 - 1)! = 120. Then, multiply by 3! for the internal arrangements.

So, same answer.

Alternatively, if I think of the total number of ways where Pierre, Rosa, and Thomas are seated together as follows:

First, consider the number of ways to seat all 8 people around the table with Pierre, Rosa, and Thomas together.

First, let's think of the number of ways to seat them in a line, then adjust for circularity.

In linear arrangements, the number of ways where Pierre, Rosa, and Thomas are together is 6! * 3! as mentioned before. But for circular arrangements, it's different.

But how does that translate?

In linear arrangements, the number is (n - k + 1)! * k!

But for circular arrangements, it's (n - k)! * k!

Wait, so in our case, n = 8, k = 3, so (8 - 3)! * 3! = 5! * 3! = 720.

Alternatively, if I think of the circular table as a line with the ends connected. So, if we fix one person's position to make it linear.

For example, fix Pierre's position to eliminate rotational symmetry. Then, we need to arrange the remaining 7 people. But with the condition that Rosa and Thomas are sitting next to Pierre? Wait, no. Wait, the condition is that Pierre, Rosa, and Thomas are all sitting together.

So, if we fix Pierre's position, then Rosa and Thomas must be seated in the two seats adjacent to Pierre? Or they can be seated anywhere as long as all three are together.

Wait, if we fix Pierre's position, then the block of Pierre, Rosa, and Thomas can be arranged in two different ways: either Rosa and Thomas are on his left and right, or they are in some order. Wait, but since it's a circular table, fixing Pierre's position, the block can be arranged in different ways.

Wait, maybe this approach is more complicated. Let me try.

Fix Pierre's seat. Now, we need to arrange the other 7 people, with the condition that Rosa and Thomas are adjacent to Pierre or somewhere else? Wait, no. The condition is that all three are sitting together. So, Pierre, Rosa, and Thomas must form a block of three consecutive seats.

Since Pierre is fixed, the block can be either to his left or to his right. Wait, but since it's a circular table, fixing Pierre's position, the block can be in two different orientations?

Wait, no. If we fix Pierre's position, then the block of three people can be arranged in two different ways: either the block is Pierre, Rosa, Thomas in that order, or Pierre, Thomas, Rosa, or other permutations. Wait, but the block can be placed in different positions.

Wait, if Pierre is fixed, then the block of three can be seated in two different positions: either the block extends clockwise from Pierre, or counter-clockwise from Pierre. But since the block is three people, if Pierre is fixed, then the block can be in positions (Pierre, seat 2, seat 3) or (seat 8, seat 1, Pierre). Wait, but seat 1 is Pierre's seat. So, if we fix Pierre at seat 1, then the block can be seats 1, 2, 3 or seats 8, 1, 2. But seat 1 is already occupied by Pierre. So, the block can be either seats 1, 2, 3 or seats 8, 1, 2. However, seat 1 is Pierre, so in the first case, the block is Pierre, seat 2, seat 3. In the second case, seat 8, Pierre, seat 2.

But in both cases, the block includes Pierre. So, the block can be in two different positions relative to Pierre's fixed seat. However, since we fixed Pierre's seat, the rest of the block can be arranged in different ways.

But maybe this is getting too complicated. Let me think.

If we fix Pierre's seat, then the number of ways to arrange the rest with Rosa and Thomas adjacent to Pierre? No, not necessarily adjacent. The block can be anywhere as long as all three are together.

Wait, if Pierre is fixed, then the block of three can be placed in different positions. For example, the block could be Pierre, Rosa, Thomas in seats 1, 2, 3; or seats 1, 8, 7; or seats 1, 2, 8? Wait, no. Wait, the block has to be three consecutive seats.

Since the table is circular, the number of ways to place a block of three consecutive seats with Pierre fixed at seat 1.

If Pierre is at seat 1, then the block can start at seat 1, 2, or 8? Wait, no. If the block is three consecutive seats, then starting at seat 1, the block would be seats 1, 2, 3. Starting at seat 8, the block would be seats 8, 1, 2. Starting at seat 7, the block would be seats 7, 8, 1. But seat 1 is already occupied by Pierre. So, if we want the block to include Pierre, then the block can be seats 8, 1, 2 or seats 1, 2, 3. So, two possible blocks that include Pierre.

But if we don't require the block to include Pierre? Wait, no. The block must include Pierre, Rosa, and Thomas. Since Pierre is fixed, the block must include Pierre. Therefore, the block can be in two different positions: either seats 1, 2, 3 or seats 8, 1, 2.

But in each case, the block includes Pierre. So, for each of these two positions, we can arrange Rosa and Thomas in the other two seats of the block.

So, for each of the two positions (block starting at seat 1 or block starting at seat 8), we can arrange Rosa and Thomas in 2! ways.

But also, the block can be arranged in different orders. Wait, if the block is seats 1, 2, 3, then Pierre is in seat 1, and Rosa and Thomas can be in seats 2 and 3 in 2! ways. Similarly, if the block is seats 8, 1, 2, then Pierre is in seat 1, and Rosa and Thomas can be in seats 8 and 2 in 2! ways.

But also, the block could be arranged in different orders. Wait, for example, if the block is seats 1, 2, 3, Pierre could be in seat 1, 2, or 3? Wait, no, because we fixed Pierre at seat 1. Therefore, Pierre must be in seat 1, and the other two seats in the block (2 and 3) are for Rosa and Thomas. Similarly, if the block is seats 8, 1, 2, then Pierre is in seat 1, and the other two seats (8 and 2) are for Rosa and Thomas.

But wait, if we fix Pierre at seat 1, then the block can be in two different positions (either extending clockwise or counter-clockwise from Pierre). However, in each case, the other two seats in the block are adjacent to Pierre. Therefore, the number of ways to arrange Rosa and Thomas in the block is 2! for each block position.

But there are two block positions, so total arrangements for the block would be 2 * 2! = 4.

Then, after placing the block, we have 5 remaining people to arrange in the remaining 5 seats. Since we fixed Pierre's position, the rest can be arranged in 5! ways.

Therefore, total number of arrangements would be 4 * 5! = 4 * 120 = 480.

Wait, but this contradicts the previous answer of 720. What's going on here?

Hmm, so according to this method, fixing Pierre's position gives 480, but earlier methods gave 720. There's a discrepancy here. Which one is correct?

Let me check.

If I fix Pierre's position, then the number of ways where Pierre, Rosa, and Thomas are all together is 2 * 2! * 5! = 480. But according to the previous methods, it's 720.

So, which is correct?

Alternatively, maybe I made a mistake in this approach. Let me think again.

If we fix Pierre's position, then the block of three can be placed in different ways. However, maybe there are more positions for the block?

Wait, if the block is three consecutive seats, and Pierre is fixed at seat 1, then the block can be:

- Seats 1, 2, 3

- Seats 8, 1, 2

- Seats 7, 8, 1

Wait, that's three different blocks that include seat 1. Wait, seat 7, 8, 1 is another block that includes seat 1. So, actually, there are three different blocks that include seat 1.

Wait, let me visualize the table. Let's label the seats 1 through 8 clockwise. If Pierre is fixed at seat 1, then the blocks of three consecutive seats that include seat 1 are:

1. Seats 1, 2, 3

2. Seats 8, 1, 2

3. Seats 7, 8, 1

So, three different blocks. Therefore, there are three positions for the block that includes seat 1.

Therefore, for each of these three blocks, we can arrange Pierre, Rosa, and Thomas in the block. However, since Pierre is fixed at seat 1, the block must include seat 1.

So, for each of these three blocks, we need to arrange Rosa and Thomas in the other two seats of the block.

For example:

- Block 1, 2, 3: Pierre is at 1, so Rosa and Thomas can be at 2 and 3 in 2! ways.

- Block 8, 1, 2: Pierre is at 1, so Rosa and Thomas can be at 8 and 2 in 2! ways.

- Block 7, 8, 1: Pierre is at 1, so Rosa and Thomas can be at 7 and 8 in 2! ways.

Therefore, total number of ways to arrange the block is 3 * 2! = 6.

Then, the remaining 5 people can be arranged in the remaining 5 seats in 5! ways.

Therefore, total number of arrangements is 6 * 5! = 6 * 120 = 720.

Ah! So earlier, I had only considered two blocks, but there are actually three blocks that include seat 1. Therefore, the correct number is 3 * 2! * 5! = 720, which matches the previous answer.

So, my mistake earlier was not considering all possible blocks that include Pierre's fixed seat.

Therefore, this method also gives 720.

Therefore, this confirms that the answer is 720.

But let me check once more with another approach.

Suppose we don't fix any positions.

The total number of circular permutations where Pierre, Rosa, and Thomas are seated together is equal to the number of ways to arrange the 8 people around the table with those three together.

As mentioned before, treating them as a single block, we have 6 entities. The number of circular arrangements is (6 - 1)! = 120. Multiply by the number of ways to arrange the block, which is 3! = 6. So, 120 * 6 = 720.

Alternatively, using the formula for circular permutations with a group together: (n - k)! * k! where n is total people, k is the size of the group.

So, (8 - 3)! * 3! = 5! * 3! = 120 * 6 = 720.

Therefore, all methods lead to 720.

But let me think if there's another way someone might approach this problem and get a different answer.

For example, someone might think: first, choose the seats for Pierre, Rosa, and Thomas. Since they must sit together, they occupy 3 consecutive seats. How many ways are there to choose 3 consecutive seats around a circular table?

In a circular table with 8 seats, the number of ways to choose a block of 3 consecutive seats is 8. Because for each seat, there is a block starting at that seat. However, since the table is circular, each block is counted once. Wait, for example, seats 1-2-3, 2-3-4,..., 8-1-2. So, there are 8 different blocks of 3 consecutive seats.

But if we consider that each block is unique, then once we choose a block, we can arrange Pierre, Rosa, and Thomas in 3! ways. Then, arrange the remaining 5 people in the remaining 5 seats in 5! ways.

But since the table is circular, are we overcounting?

Wait, if we consider that choosing different blocks might lead to equivalent arrangements due to rotation.

But if we fix the block, then we have to consider that rotations are considered the same.

Wait, if we first count the number of ways as follows:

Number of ways = (number of blocks) * (arrangements within block) * (arrangements of others).

But since rotations are considered the same, how does that affect?

If we count the number of blocks as 8, then multiply by 3! * 5!, we would get 8 * 6 * 120 = 5760. But this is way larger than the total number of circular permutations, which is 7! = 5040.

Therefore, clearly, this approach is wrong.

So, why is that? Because when we count the number of blocks as 8, we are considering different rotations as different blocks, but in reality, in circular permutations, rotations are considered the same. Therefore, we need to adjust for that.

Therefore, if we use this method, we have to divide by the number of rotations.

Wait, but how?

Alternatively, if we consider that each unique circular arrangement is counted multiple times in the block count.

For example, each unique arrangement where the block is seated somewhere is counted once for each rotation.

But since the block can be rotated, but the entire arrangement is considered the same.

This seems complicated.

Alternatively, since we know that the total number of circular arrangements is 7! = 5040, and the number of arrangements where the block is seated together is 720, then 720 / 5040 = 1/7. Which makes sense, because there are 8 possible positions for the block, but since rotations are considered the same, each unique arrangement is counted 8 times? Wait, no.

Wait, if we fix the block, then the number of unique arrangements is (number of blocks) * ...

But this seems messy.

Alternatively, since we know that the answer is 720, and the total number of circular permutations is 5040, then the probability that a random seating has the three people together is 720 / 5040 = 1/7.

But is that correct?

Alternatively, the number of ways to seat the three people together is (number of ways to seat them as a block) = (n - 1)! * k! / n? Wait, not sure.

But since we have already confirmed with multiple methods that the answer is 720, and the alternative approach of counting blocks led to overcounting, I think the correct answer is 720.

Therefore, after all these verifications, I think the answer is 720.

But just to make sure, let me think of a smaller case. Let's take n=3, and the group size k=3. Then, the number of circular arrangements where all three are together is 1 (since they are all together). According to the formula (n - k)! * k! = (3 - 3)! * 3! = 1 * 6 = 6. But that's wrong because in circular permutations, there's only 1 way to arrange 3 people.

Wait, this is a problem.

Wait, hold on. If n=3, and k=3, then according to the formula (n - k)! * k! = 0! * 3! = 1 * 6 = 6. But the actual number of circular permutations is (3 - 1)! = 2. So, clearly, the formula is wrong here.

Wait, so my previous reasoning must have a mistake.

Wait, so what's the problem here?

In the case of n=3, k=3, the number of circular arrangements where all three are together is 1 (since they are all together). But according to the formula (n - k)! * k! = 1 * 6 = 6. Which is wrong.

So, there's a mistake in my previous reasoning.

Wait, so where is the error?

Earlier, I thought that the formula is (n - k)! * k! for circular permutations with a group of k people together. But in this case, it's not.

So, let's think again.

For n=3, k=3. How many circular arrangements where all three are together? Well, since they are all together, it's just the number of circular permutations of 3 people, which is (3 - 1)! = 2. But according to the formula (n - k)! * k! = 1 * 6 = 6. Which is wrong.

Therefore, my previous formula is incorrect.

So, what's the correct formula?

Let me think again.

When dealing with circular permutations where a group of k people must sit together, the correct approach is to treat the group as a single entity. Then, the number of entities is (n - k + 1). Wait, no.

Wait, if we have n people, and we treat a group of k as a single entity, then the number of entities is (n - k + 1)? No, it's (n - k + 1) if we are considering linear arrangements.

Wait, no. If you have n people and you group k of them together, you have (n - k + 1) entities? No, that's for linear arrangements.

Wait, in linear arrangements, the number of ways to arrange n people with a group of k together is (n - k + 1)! * k!

But for circular arrangements, it's different.

Let me think again.

In circular arrangements, when we have a group of k people that must sit together, we can consider them as a single entity. Then, the number of entities is (n - k + 1)? No, it's (n - k + 1) if we are considering linear arrangements.

Wait, no. If we have n people, and we group k of them into a single entity, then the number of entities is (n - k + 1)? No, it's (n - k + 1) only if we are considering the group as a single entity and the rest as individuals. Wait, no.

Actually, if you have n people and you group k of them into a single entity, then the number of entities is (n - k) + 1 = n - k + 1.

But in circular permutations, the number of ways to arrange these entities is (number of entities - 1)!

Therefore, total number of arrangements is (n - k + 1 - 1)! * k! = (n - k)! * k!

But in the case of n=3, k=3, this gives (3 - 3)! * 3! = 1 * 6 = 6, which is wrong.

But the correct number is 2.

So, clearly, this formula is not correct.

Therefore, my previous reasoning was flawed.

So, what is the correct formula?

Let me think again.

In the case of n=3, k=3. If we treat the group as a single entity, then we have 1 entity. The number of circular arrangements of 1 entity is (1 - 1)! = 0! = 1. Multiply by the internal arrangements of the group, which is 3! = 6. So total is 1 * 6 = 6. But the actual number of circular arrangements is 2.

So, there's a contradiction here.

Wait, why is that?

Because when we treat the group as a single entity, we are assuming that the group can be placed anywhere, but in reality, when the group is the entire set, the circular arrangement is just the number of circular permutations of the entire group.

But in this case, the formula gives 6, but the correct answer is 2.

So, the problem is that when the group is the entire set, the formula overcounts.

Wait, but why?

Because when we treat the group as a single entity, we are considering that the group can be arranged in k! ways, but in reality, in circular permutations, the entire group's arrangement is already considered.

Wait, maybe the formula is correct, but the interpretation is wrong.

Wait, if we have n=3, and we treat the group of 3 as a single entity, then the number of circular arrangements is (1 - 1)! * 3! = 1 * 6 = 6. But in reality, the number of circular arrangements of 3 people is 2.

So, why is there a discrepancy?

Because when we treat the group as a single entity, we are not accounting for the fact that the entire group is the whole table.

Wait, in other words, when we have a single entity, the number of circular arrangements is 1 (since there's only one entity), but we multiply by the internal arrangements. However, in reality, the internal arrangements are already considered in the circular permutations.

Wait, no. In circular permutations, when you have a single entity, it's just one way to arrange it, but the internal arrangements are separate.

But in reality, arranging the single entity (which is the entire group) around the table is just one way, but the internal arrangements are the permutations of the group. However, in circular permutations, the internal arrangements of the group would be considered as linear arrangements, but since the entire table is circular, does that affect it?

Wait, for example, if you have three people around a table, the number of circular permutations is 2. However, if you consider them as a single entity, then the number of circular arrangements is 1 (for the entity), multiplied by 3! (internal arrangements). But 1 * 6 = 6, which is not equal to 2.

Therefore, there's a problem here.

So, what is the correct way to think about this?

Perhaps the formula (n - k)! * k! is not correct for circular permutations.

Let me think again.

In the case of n=3, k=3, the correct answer is 2.

If we use the method of fixing the block's position.

Fix one person's position to eliminate rotational symmetry. Let's fix Pierre at seat 1. Then, the other two people, Rosa and Thomas, can be arranged in 2! ways. So total is 2. Which matches the correct answer.

But according to the formula (n - k)! * k! = 1 * 6 = 6, which is wrong.

Therefore, the formula is incorrect.

So, what is the correct formula?

Let me think again.

When we have a group of k people that must sit together, we can treat them as a single entity. Then, the number of entities is (n - k + 1)? No, it's (n - k + 1) if we are considering linear arrangements.

Wait, no. If we have n people and we group k of them into a single entity, then the number of entities is (n - k) + 1 = n - k + 1.

But in circular arrangements, the number of ways to arrange these entities is (number of entities - 1)!

Therefore, total number of arrangements is (n - k + 1 - 1)! * k! = (n - k)! * k!

But this gives the wrong answer for n=3, k=3.

So, why?

Because when the group is the entire set, the formula counts the internal arrangements as k! but in reality, the circular permutations already account for rotational symmetry.

Wait, maybe the formula is correct, but when we fix the group as a single entity, we need to adjust for rotational symmetry.

Wait, if we have n=3, k=3, then the number of entities is 1. The number of circular arrangements of 1 entity is 1 (since there's only one way to arrange one entity around a table). Multiply by the internal arrangements of the entity, which is 3! = 6. But in reality, the number of circular arrangements is 2.

So, there's a contradiction.

Wait, but if we think of the internal arrangements as linear arrangements, then maybe we need to divide by something?

Wait, no. In circular permutations, when you have a single entity, the internal arrangements are linear?

Wait, no. If the entire group is the single entity, then the internal arrangements are the circular permutations of the group?

Wait, no. If the entire group is considered as a single entity, then the internal arrangements are the permutations of the group, but since the entire table is circular, the internal arrangements are actually linear?

This is confusing.

Let me think of another example. Suppose n=4, k=2. How many circular arrangements where two specific people are together?

Using the formula (n - k)! * k! = (4 - 2)! * 2! = 2! * 2! = 4.

But let's count manually.

Total circular permutations of 4 people: (4 - 1)! = 6.

Number of circular arrangements where two specific people are together:

Treat the two people as a block. Then, we have 3 entities: the block and the other two people. Number of circular arrangements: (3 - 1)! = 2. Multiply by internal arrangements of the block: 2! = 2. Total: 2 * 2 = 4. Which matches the formula.

But let's count manually.

Fix one person, say Pierre, to eliminate rotational symmetry. Then, the other person (Rosa) can be either to his left or right. So, 2 choices. Then, the remaining two people can be arranged in 2! ways. So total is 2 * 2 = 4. Which matches.

So, in this case, the formula works.

But in the case of n=3, k=3, the formula gives 6, but the correct answer is 2.

Wait, why is that?

If we use the same method: fix one person, say Pierre. Then, the other two people (Rosa and Thomas) can be arranged in 2! ways. So total is 2. Which is correct.

But according to the formula (n - k)! * k! = 1 * 6 = 6.

So, why is there a discrepancy?

Ah! Because when we treat the group as a single entity, and calculate (number of entities - 1)! * k! = (1 - 1)! * 3! = 1 * 6 = 6, but in reality, when we fix one person, we are effectively accounting for rotational symmetry, and the formula should be adjusted.

Wait, so maybe the formula is not (n - k)! * k! but something else.

Wait, in the case of n=3, k=3, if we treat the group as a single entity, then the number of circular arrangements is (number of entities - 1)! = 0! = 1, and then multiply by the internal arrangements of the group, which is 3! = 6. But in reality, the number of circular arrangements is 2.

So, why is there a difference?

Because when we treat the group as a single entity, we are not accounting for the fact that the entire group is the whole table.

Wait, if we have a single entity, which is the entire group, then arranging it around the table is just 1 way, but the internal arrangements are 3! = 6. However, in reality, the circular permutations of the entire group is (3 - 1)! = 2.

So, the formula is overcounting by a factor of 3.

Wait, why?

Because when we treat the group as a single entity, we are considering the internal arrangements as linear, but in reality, they should be circular.

Ah! So, the mistake is that when we treat the group as a single entity, we should consider the internal arrangements as circular permutations, not linear.

Wait, but the group is a block that is part of the circular table.

Wait, this is getting too tangled.

Let me think again.

If we have a group of k people that must sit together, and we treat them as a single entity, then the number of ways to arrange the entities around the table is (number of entities - 1)!

But the internal arrangements of the group: if the group is a block, then their internal arrangements are linear, because the block is part of the circular table.

Wait, no. If the block is part of the circular table, then the internal arrangements of the block are linear? Or are they circular?

Actually, the internal arrangements of the block are linear because the block is a consecutive sequence. For example, if the block is three people sitting together, their arrangement is linear relative to the rest of the table.

Wait, but in reality, the entire table is circular, so the block's internal arrangement is a linear arrangement, because the block is a consecutive segment.

Therefore, the internal arrangements should be k!

But in the case of n=3, k=3, the internal arrangements would be 3! = 6, but the actual number of circular permutations is 2.

So, why is there a discrepancy?

Because when the entire group is the table, the internal arrangements are actually circular.

Wait, if the entire group is the table, then the internal arrangements are circular. So, in that case, the number of internal arrangements should be (k - 1)! instead of k!

Ah! So, the formula should be:

If we have a group of k people that must sit together, treat them as a single entity. The number of circular arrangements is (number of entities - 1)! * (k - 1)!

Wait, let's test this.

For n=3, k=3: number of entities = 1. So, (1 - 1)! * (3 - 1)! = 1 * 2 = 2, which is correct.

For n=4, k=2: number of entities = 2 (the block and the other two people). Wait, no. If n=4, k=2, then number of entities is (4 - 2) + 1 = 3? Wait, no.

Wait, if we have n=4 people, and we group k=2 into a block, then the number of entities is 3: the block and the other two individuals.

So, number of circular arrangements is (3 - 1)! * (2 - 1)! = 2! * 1! = 2 * 1 = 2. But earlier, we saw that the correct number is 4.

Wait, this is conflicting.

Wait, earlier, for n=4, k=2, the correct number of circular arrangements where the two people are together is 4. But according to this new formula, it would be (number of entities - 1)! * (k - 1)! = (3 - 1)! * (2 - 1)! = 2! * 1! = 2. Which is wrong.

So, this suggests that this formula is also incorrect.

This is getting too confusing. Let me try to find a general formula.

Let me refer back to standard combinatorial results.

In circular permutations, the number of ways to arrange n people with a specific group of k people together is (n - k)! * k!

Wait, but this gave the wrong answer for n=3, k=3.

But in the case of n=4, k=2, it gave 2! * 2! = 4, which is correct.

Wait, for n=3, k=3, it gives 0! * 3! = 1 * 6 = 6, which is wrong.

But why is it correct for n=4, k=2 but not for n=3, k=3?

Maybe because when the group is the entire set, the formula overcounts.

Wait, but in the case of n=3, k=3, the formula gives 6, but the correct answer is 2.

But if we think of the formula as (n - k)! * k! = (3 - 3)! * 3! = 1 * 6 = 6, but the actual number of circular permutations is 2.

So, the formula is overcounting by a factor of 3.

Wait, but why?

Because when we treat the group as a single entity, we are considering the internal arrangements as linear, but in reality, when the group is the entire set, the internal arrangements should be circular.

Wait, but in the case of n=4, k=2, the group is not the entire set, so the internal arrangements are linear.

So, perhaps the formula is correct when the group is not the entire set, but when the group is the entire set, we need to adjust.

But this seems inconsistent.

Alternatively, maybe the formula is correct, and the problem is in the interpretation of the question.

Wait, in the original problem, the question says: "Two seatings are considered the same if one is a rotation of the other."

So, in the case of n=3, k=3, the number of distinct seatings is 2, but according to the formula (n - k)! * k! = 6, which suggests that there are 6 distinct seatings, which is wrong.

But why?

Because when we use the formula (n - k)! * k!, we are assuming that the group is treated as a single entity, and the rest are individuals, and then arranging them around the table.

But in the case where the group is the entire set, this would mean arranging 1 entity (the group) around the table, which is 1 way, multiplied by k! = 6, giving 6. But in reality, the number of distinct seatings is 2.

So, the formula is wrong.

But why does it work for n=4, k=2?

Let me think. For n=4, k=2:

Using the formula: (4 - 2)! * 2! = 2! * 2! = 4.

Which matches the correct answer.

But if we use the same logic as for n=3, k=3, why is it different?

Maybe because when the group is not the entire set, the formula works, but when it is, it doesn't.

This suggests that the formula is not universally applicable.

Alternatively, perhaps the formula is correct, but the interpretation of the problem is different.

Wait, in the case of n=3, k=3, if we consider the formula (n - k)! * k! = 6, but the actual number of distinct seatings is 2, this suggests that the formula counts each seating multiple times.

Wait, how many times is each seating counted in the formula?

If we consider that each seating is counted k times, because the group can be rotated.

Wait, for n=3, k=3, each seating is counted 3 times in the formula.

Because the formula counts all possible rotations of the group.

But since rotations are considered the same, we need to divide by the number of rotations.

Wait, but in the formula, we already accounted for rotational symmetry by using (number of entities - 1)!

Wait, this is getting too tangled.

Maybe I should refer to standard references.

Wait, according to standard combinatorics, the number of ways to arrange n people around a circular table with a specific group of k people together is (n - k)! * k!

But this seems to conflict with the n=3, k=3 case.

Wait, let me check an external source.

Wait, I can't access external sources, but I can think.

In the case of circular permutations, when we have a group of k people that must sit together, the standard approach is to treat the group as a single entity, leading to (n - k + 1 - 1)! * k! = (n - k)! * k!

Wait, but this is the same as before.

But in the case of n=3, k=3, this gives 1 * 6 = 6, but the correct answer is 2.

So, what is the issue?

Ah! I think the issue is that when the group is the entire set, the formula counts the number of linear arrangements, not circular.

Wait, no.

Wait, let's think of it this way: when you treat the group as a single entity, you are effectively converting the circular permutation problem into a linear one.

For example, if you have a circular table with n people, and you fix one person's position to make it linear, then the number of arrangements is (n - 1)!

But if you have a group of k people, you can fix the group's position to make it linear.

Wait, this is getting too abstract.

Let me try to think of the n=3, k=3 case.

If we have 3 people around a table, and we want to count the number of circular arrangements where all three are together. Well, they are always together, so it's just the number of circular permutations, which is 2.

But according to the formula (n - k)! * k! = 1 * 6 = 6.

So, why is there a discrepancy?

Because when we treat the group as a single entity, we are not accounting for the fact that the group's internal arrangements are also circular.

Wait, if the group is a single entity, then its internal arrangements are circular?

No, the group is a block of people sitting together, so their internal arrangement is linear.

Wait, but in the case of the entire group, the block is the whole table, so their internal arrangement is circular.

Therefore, in this case, the internal arrangements should be (k - 1)! instead of k!

So, if we adjust the formula to (n - k)! * (k - 1)!

For n=3, k=3: (3 - 3)! * (3 - 1)! = 1 * 2 = 2, which is correct.

For n=4, k=2: (4 - 2)! * (2 - 1)! = 2! * 1! = 2 * 1 = 2, which is wrong, because earlier we saw that the correct number is 4.

Wait, now it's conflicting again.

This suggests that the formula is not consistent.

This is very confusing.

Let me try to think of the general case.

When we have a group of k people that must sit together, we can treat them as a single entity. The number of ways to arrange the entities around the table is (number of entities - 1)!

The number of ways to arrange the group internally is k! if the group is treated as a linear block, or (k - 1)! if the group is treated as a circular block.

But in the context of the entire table being circular, the group's internal arrangement is linear, because the group is part of the circular table.

Wait, for example, if you have a block of k people sitting together, their arrangement is linear relative to the rest of the table.

Therefore, the internal arrangements should be k!

But in the case where the entire table is the group, the internal arrangement is circular, so it should be (k - 1)!

This suggests that the formula depends on whether the group is the entire table or not.

This seems too complicated.

Perhaps the correct approach is to use the method of fixing a position to eliminate rotational symmetry.

For example, in the original problem, with n=8, k=3.

Fix one person's position, say Pierre. Then, the number of ways to arrange the rest with Rosa and Thomas sitting together.

Wait, but this is what I did earlier, and it gave 720.

But in the case of n=3, k=3, fixing Pierre's position, the number of ways is 2, which is correct.

So, perhaps the correct formula is:

Number of circular arrangements = (number of ways to arrange the rest with the group together) = (n - 1)! * (k / n) ?

No, not sure.

Alternatively, the number of ways is (n - 1)! * (k / n) ?

Wait, for n=3, k=3: (3 - 1)! * (3 / 3) = 2 * 1 = 2, which is correct.

For n=4, k=2: (4 - 1)! * (2 / 4) = 6 * 0.5 = 3, which is wrong, because the correct number is 4.

Hmm, no.

Alternatively, the number of ways is (n - 1)! * (k! / n) ?

For n=3, k=3: 2! * (6 / 3) = 2 * 2 = 4, which is wrong.

This is not working.

Let me go back to the original problem.

Since in the original problem, using the method of fixing Pierre's position gave 720, and other methods also gave 720, and in the n=3, k=3 case, fixing one person's position gives the correct answer, perhaps the formula is:

Number of circular arrangements = (n - 1)! * (k! / n) ?

Wait, no.

Wait, in the original problem, if we fix Pierre's position, then the number of ways is:

Number of ways to arrange the rest with Rosa and Thomas together.

Which, as we calculated earlier, is 3 * 2! * 5! = 720.

But (n - 1)! is 7! = 5040.

So, 720 is 5040 * (720 / 5040) = 5040 * (1/7).

So, the probability is 1/7.

But why 1/7?

Because there are 8 possible positions for the block of 3 people, but since rotations are considered the same, each unique arrangement is counted 8 times?

Wait, no.

Alternatively, the number of ways to choose the block is 8, but each unique arrangement is counted 3 times (since the block can start at any of the 3 positions).

Wait, this is getting too tangled.

But since in the original problem, all methods gave 720, and in the n=3, k=3 case, fixing one person's position gives the correct answer, I think that the correct answer for the original problem is 720.

Therefore, despite the confusion with the n=3, k=3 case, I think that the original answer of 720 is correct.

Perhaps the formula (n - k)! * k! is correct when the group is not the entire set, and in the case where the group is the entire set, it's a special case.

But since in the original problem, the group is not the entire set, and all methods give 720, I will go with 720 as the answer.

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

**Final Answer**
The number of ways is \boxed{720}.

To determine the number of ways 8 people can sit around a round table such that **Pierre, Rosa, and Thomas all sit together**, we need to carefully account for both the **circular nature of the table** and the **restriction that three specific people must be adjacent**.

---

### Step 1: Understanding Circular Permutations

In a **circular arrangement**, the number of distinct ways to arrange $ n $ people is:

$$
(n - 1)!
$$

This is because rotations of the same arrangement are considered identical. For example, the arrangement ABCDE is the same as BCDEA in a circular table.

---

### Step 2: Treating the Group as a Single Entity

Since **Pierre, Rosa, and Thomas** must sit together, we can treat them as a **single block** or **entity**. This reduces the problem to arranging:

- 1 block (Pierre, Rosa, Thomas)
- 5 other individuals

So, we now have **6 entities** to arrange around the table.

---

### Step 3: Arranging the 6 Entities

The number of ways to arrange 6 entities in a circle is:

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

---

### Step 4: Arranging the People Within the Block

Within the block of 3 people (Pierre, Rosa, and Thomas), they can be arranged among themselves in:

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

---

### Step 5: Total Number of Arrangements

Now, multiply the number of ways to arrange the 6 entities by the number of internal arrangements within the block:

$$
5! \times 3! = 120 \times 6 = 720
$$

---

### 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 $3! \times 5! = \boxed{720}$.