<Question>
In how many ways can 8 people sit around a round table if 3 of the people -- Pierre, Rosa, and Thomas -- all want to sit together?  (Two seatings are considered the same if one is a rotation of the other.)
</Question>

<think>
<Plan_1>
Okay, so I need to figure out how many ways 8 people can sit around a round table, but Pierre, Rosa, and Thomas must all sit together. Hmm, circular permutations can be tricky, but let me try to break it down step by step.

<prompt_1.1> Identify the problem as a circular permutation problem and recall the formula for arranging people around a circular table. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, I remember that when arranging people around a circular table, the number of arrangements is (n-1)! for n people because rotations are considered the same. But here, there's a condition that Pierre, Rosa, and Thomas must sit together. So, maybe I can treat them as a single unit or "block"?

Let me think. If we have 8 people but 3 of them must be together, we can consider them as one entity. So instead of 8 individuals, we have 6 other people plus this block, making it 7 total units. But since it's a circular table, the formula would be (7-1)! for arranging these units.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
That would be 6! .

But wait, within the block, Pierre, Rosa, and Thomas can be arranged among themselves in different ways. Since there are 3 people in the block, they can be arranged in 3! ways.

So, maybe the total number of arrangements is 6! * 3! .

Let me check if that's correct.

But hold on, is there anything else I need to consider?

<prompt_2.1> Verify the calculation by considering the arrangement of the block and the application of the formula for circular permutations. </prompt_2.1>

Wait, but let me make sure I didn't miss anything. For example, sometimes in circular permutations, fixing a position can help avoid overcounting. Let me try another approach to verify.

<prompt_2.2> Verify the result using an alternative method by linearizing the problem and fixing a position. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let me think again. If we fix the block as a single entity, then the 7 entities (block + 6 others) are arranged around the table. Since it's circular, the number of arrangements is indeed (7-1)! = 6! . Then, multiplying by 3! for the internal arrangements of the block. So, yes, 6! * 3! .
</execution_2.1>

<execution_2.2>
Suppose we treat Pierre, Rosa, and Thomas as a single entity. Then, as I thought earlier, we have 6 entities around the table. The number of ways to arrange 6 entities around a circular table is (6-1)! = 5! . Wait, but why did I say 6 instead of 7? Oh, right, because the block is considered one entity, so total entities are 6 (the block plus 5 other people). Therefore, circular permutations would be (6-1)! = 5! .

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

Hmm, now there's a discrepancy here. The first method gave me 6! * 3! and this method gives 5! * 3! . Which one is correct?

Wait, maybe I made a mistake in the first approach. Let me clarify. If we have 8 seats around a table, and we want to arrange 8 people with Pierre, Rosa, and Thomas together. If we treat them as a single block, then we have 6 other people plus 1 block, totaling 7 units. But when arranging 7 units around a circular table, the formula is (7-1)! = 6! .
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
So which is correct?

Wait, maybe the confusion is between linear and circular arrangements.

<prompt_3.1> Resolve the discrepancy between the two methods by carefully considering the number of units and their arrangement. </prompt_3.1>

But why does the second approach give 5! * 3! ? Let me check that again.

<prompt_3.2> Re-evaluate the second approach to understand why it gives a different result. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
In linear arrangements, the number is n! for n units, but in circular, it's (n-1)! because rotations are equivalent. So if we have 7 units, it's 6! .

But in the problem, the total number of people is 8, but when we form a block of 3, we have 6 units (7-1? ) Wait, no. Let me recount. If we have 8 people, and 3 are in a block, then the number of units is 8 - 3 + 1 = 6. So 6 units. Therefore, arranging 6 units around a circular table is (6-1)! = 5! . Then multiply by 3! for the internal arrangements. So that would be 5! * 3! .

But wait, that contradicts my first approach. So which is correct?
</execution_3.1>

<execution_3.2>
If we fix the block in a certain position, then the remaining 5 people are arranged linearly? No, even if we fix the block, the table is still circular. So actually, if the block is considered as a single entity, then the number of entities is 6 (block + 5 others). Wait, no, original total people are 8. If we make a block of 3, then the number of entities is 6 (3 in the block and 5 others). Wait, 3 in the block plus 5 others is 8 people total. So, yes, 6 entities. Therefore, circular permutations would be (6-1)! = 5! .

But this contradicts the first method. So which is correct?

Wait, perhaps the key is whether the block is considered as a single entity. Let me clarify. In circular permutations, when we have a block, the formula is (number of entities -1)! * permutations within the block. So if we have k entities, it's (k-1)! * m! where m! is the permutations within the block.

So in this case, if we have 6 entities (block + 5 others), then it's (6-1)! * 3! = 5! * 3! .
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Hold on, maybe the error is in the first approach where I considered 7 units. Let me check again.

<prompt_4.1> Clarify the number of units and confirm the correct approach for arranging the block. </prompt_4.1>

Based on execution_3.2:
But why does this differ from the first approach?

<prompt_4.2> Reconcile the two methods to understand why they give different results. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Original problem: 8 people, 3 must sit together. So, total people: 8. If we make a block of 3, then the number of "units" to arrange is 8 - 3 + 1 = 6. So 6 units. Therefore, circular permutations: (6-1)! = 5! . Then, multiply by 3! for the block's internal arrangements. So total is 5! * 3! .

But in my first approach, I considered 7 units, which was incorrect.
</execution_4.1>

<execution_4.2>
Wait, perhaps the first approach was incorrect because the number of units is actually 6, not 7. Let me recount: original people are 8. If Pierre, Rosa, and Thomas are a block, then instead of 3 individuals, we have 1 block. Therefore, total units to arrange are 8 - 3 + 1 = 6. So 6 units. Therefore, circular permutations are (6-1)! = 5! , and multiplied by 3! for the block. So total is 5! * 3! .

But then, why did I initially think of 7 units? Because maybe I confused the total number of entities. So the correct number is 6 units, leading to 5! * 3! .

But let me check with a smaller example to verify. Suppose there are 4 people, A, B, C, D. If we want to arrange them around a table with A and B must sit together. Then, treating A and B as a block, we have 3 units: AB, C, D. The number of circular arrangements is (3-1)! = 2! , and within the AB block, 2! arrangements. So total 2! * 2! = 4. Let's list them:

- AB, C, D: arrangements around the table would be AB-C-D, AB-D-C, C-AB-D, C-D-AB, D-AB-C, D-C-AB. Wait, but in circular tables, rotations are the same. So starting with AB, C, D: AB can be AB or BA. If we fix AB in one position, then C and D can be arranged in two ways: C-D or D-C. But since it's a circle, once AB is fixed, rotating the table would move C and D. Wait, maybe in the linear case, but in circular, fixing one position, the number is (n-1)! .

Wait, in the small example with 4 people and A and B together, the formula gives 2! * 2! = 4. Let's see if that's correct. The possible distinct arrangements are:

1. AB-C-D
2. AB-D-C
3. BA-C-D
4. BA-D-C

But in a circular table, AB-C-D is the same as C-D-AB because rotating the table would make it equivalent. Wait, no. If we fix the position of AB, then C and D can be arranged in two ways. But since it's a circle, if we don't fix any position, then the total number is indeed (3-1)! * 2! = 2! * 2! = 4. Let's confirm:

The distinct arrangements considering rotations as the same are:

1. AB, C, D
2. AB, D, C
3. BA, C, D
4. BA, D, C

Yes, that's 4, which matches the formula. So in this case, the formula works. Therefore, applying the same logic to the original problem: 6 units (block + 5 others) arranged circularly, so (6-1)! = 5! , multiplied by 3! for the block. So total arrangements: 5! * 3! = 120 * 6 = 720.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
But wait, let me check another way. Suppose we "break" the circular table by fixing the position of the block. If we fix the block in a certain position, then the remaining 5 people can be arranged linearly.

<prompt_5.1> Verify the solution by considering the fixing of the block's position and calculating the arrangements. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
But since the table is circular, fixing the block's position accounts for the rotational symmetry. Then, the remaining 5 people can be arranged in 5! ways. Then, within the block, 3! ways. So total is 5! * 3! = 120 * 6 = 720. Which matches the previous result.

Therefore, the answer is 720. So the initial confusion was because of miscounting the number of units. The key is that when you form a block of 3 from 8 people, you have 6 units total, leading to (6-1)! arrangements, which is 5! , multiplied by 3! for the block. So the correct answer is 5! * 3! = 720.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the number of ways 8 people can sit around a round table with Pierre, Rosa, and Thomas sitting together is $\boxed{720}$.
</Final_answer>
</think>