<Question>
A Senate committee has 5 Democrats, 5 Republicans, and 1 Independent.  In how many ways can they sit around a circular table if all the members of each party all sit next to each other?  (Two seatings are considered equivalent if one is a rotation of the other.)
</Question>

<think>
<Plan_1>
Okay, so I have this problem here: There's a Senate committee with 5 Democrats, 5 Republicans, and 1 Independent. They need to sit around a circular table, but all members of each party must sit next to each other. The question is asking how many different ways they can arrange themselves around the table, considering that two seatings are equivalent if one is a rotation of the other.

Hmm, let me try to figure this out step by step.

<prompt_1.1> Identify the key elements of the problem and set up the initial framework for solving it. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, since it's a circular table, the arrangement is rotationally invariant. That means the starting point doesn't matter, which is different from linear arrangements. For circular permutations, the formula for arranging n distinct objects is (n-1)! because rotating the entire group doesn't create a new arrangement. But here, there's a catch: the parties must sit together. So, it's not just arranging individuals around a table, but arranging groups first, and then within each group.

Wait, so there are three parties: Democrats, Republicans, and Independent. Each party has to sit together. So, first, we need to arrange the three parties around the table, and then within each party, arrange the members.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But since the table is circular, arranging the three parties around the table would have some considerations. Normally, for n objects, the number of circular permutations is (n-1)! , but here the "objects" are the parties. However, the parties are not distinguishable in terms of rotation?

<prompt_2.1> Calculate the number of ways to arrange the three parties around the circular table, considering rotational equivalence. </prompt_2.1>

But hold on, the problem says all members of each party must sit next to each other. So, each party is a block. So, we have three blocks: Democrat, Republican, Independent. Each block will occupy consecutive seats around the table.

<prompt_2.2> Determine the number of ways to arrange the party blocks around the table, considering rotational equivalence. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, no, the parties are different: Democrats, Republicans, and Independent. So, if we consider the table as a circle, the arrangement of the parties would be affected by rotation. For example, if we rotate the entire arrangement so that the Democrat block moves to where Republican was, that would be a different arrangement. Wait, no. Wait, no, because the problem says two seatings are equivalent if one is a rotation of the other. So, if the parties are arranged as Democrat, Republican, Independent in a certain order, and then rotated, that would be considered the same arrangement. Therefore, the number of distinct arrangements of the parties would be (3-1)! = 2! = 2 ways.
</execution_2.1>

<execution_2.2>
Since the table is circular, the number of ways to arrange the three blocks would be (3-1)! = 2! = 2. But wait, is that correct? Let me think.

If we have three blocks around a circular table, the number of distinct arrangements is indeed (3-1)! because rotating the entire arrangement doesn't create a new one. So, for example, if the blocks are ordered Democrat-Republican-Independent clockwise, that's the same as Republican-Independent-Democrat clockwise if you rotate the table. So, the number of distinct arrangements of the blocks is 2.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
But now, within each block, the members can be arranged among themselves. For the Democrats, there are 5 members, so they can be arranged in 5! ways. Similarly, Republicans can be arranged in 5! ways. The Independent is just one person, so there's only 1 way to arrange them.

<prompt_3.1> Calculate the total number of arrangements by considering the arrangements of the party blocks and their members. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Therefore, multiplying these together, the total number of arrangements should be 2 (for the blocks) * 5! (Democrats) * 5! (Republicans) * 1 (Independent). So, 2 * 5! * 5! .

Wait, but hold on. Let me confirm this. Let's break it down.

First, arranging the blocks around the table: since it's a circular table, the number of ways to arrange the three blocks is (3-1)! = 2. Then, for each block, we arrange the members within the block. Since each party's members are sitting together, the Democrat block has 5 people, so 5! arrangements, Republicans 5! , and Independent 1! (which is 1). So, multiplying these together: 2 * 5! * 5! * 1! = 2 * 5! * 5! .

Therefore, the total number of ways is 2*(5! )^2.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let me check if this makes sense.

<prompt_4.1> Verify the calculation of the total number of arrangements. </prompt_4.1>

But wait, let me think again. Is there a possibility that the Independent member's position affects the count? Since the Independent is a single person, their position is fixed once the blocks are arranged. For example, if the blocks are Democrat, Republican, Independent, then the Independent is always the third block. So, their position is determined by the arrangement of the blocks.

<prompt_4.2> Consider the effect of the Independent member's position on the arrangement. </prompt_4.2>

But let me verify with a smaller example. Suppose there are 1 Democrat, 1 Republican, and 1 Independent. Then, the number of arrangements should be (3-1)! = 2 for the blocks. Then, each block has 1 person, so 1! each. So total arrangements 2*1*1*1=2, which is correct. If you have the blocks arranged as D-R-I and D-I-R (considering circular rotations), but since rotating D-R-I would give R-I-D, which is considered the same. Wait, no, in the case of 1 each, the two arrangements are D-R-I and D-I-R, but since it's circular, D-R-I and R-I-D and I-D-R are the same. Wait, no. Wait, if all parties have one person, arranging them around a circular table with three seats. The number of distinct arrangements is (3-1)! = 2. So, yes, that's correct.

Now, if we have two Democrats, two Republicans, and one Independent. Then, the blocks are D (2), R (2), I (1). The number of ways to arrange the blocks is (3-1)! = 2. Then, within D: 2! , within R: 2! , within I: 1! . So total arrangements: 2 * 2! * 2! * 1! = 2*2*2*1=8.

Let me see if that's correct. Suppose the blocks are arranged as D-R-I. Then, the D block can be arranged in 2! ways, R block in 2! ways, and I is fixed. So, for each block arrangement, there are 2*2=4 arrangements. Since there are two block arrangements (D-R-I and D-I-R, but considering circular rotations), so total 2*4=8. That seems correct.

Therefore, applying the same logic to the original problem, with 5 Democrats, 5 Republicans, and 1 Independent. The blocks are D (5), R (5), I (1). The number of ways to arrange the blocks is (3-1)! = 2. Then, within D: 5! , within R: 5! , within I: 1! . So total arrangements: 2 * 5! * 5! * 1! = 2*(5! )^2.

Wait, but hold on. Wait, when we have the blocks arranged around the table, the size of the blocks might affect the total number of seats. Wait, but the problem doesn't specify the total number of seats. However, since all members of each party must sit next to each other, the total number of seats must be equal to the sum of the sizes of the blocks. So, Democrats:5, Republicans:5, Independent:1. So total seats: 5+5+1=11. So, the circular table must have 11 seats. Therefore, arranging 11 people around a circular table, but with the constraint that the 5 Democrats are together, the 5 Republicans are together, and the Independent is alone.

But in circular permutations, when we have blocks, the formula is a bit different. Wait, maybe my initial approach is correct. Let me recall the formula for arranging objects in a circle with blocks.

<prompt_4.3> Re-evaluate the approach for arranging blocks around a circular table. </prompt_4.3>
-----

Therefore, the answer is 2*(5! )^2. Let me compute that value to check.

<prompt_4.4> Recalculate and verify the total number of arrangements. </prompt_4.4>
</Plan_4>

<Execution_4>
<execution_4.1>
Yes, arranging the blocks: (3-1)! = 2. Then, each party's internal arrangements: 5! for Democrats, 5! for Republicans, and 1! for the Independent. So, 2 * 5! * 5! * 1! = 2*(5! )^2. That seems right.

But let me check if there's any step I missed. For instance, does the circular arrangement affect the internal permutations? For example, if the table has 11 seats, and the blocks are of sizes 5,5,1. Then, once the blocks are arranged around the table, the positions are fixed relative to each other. Then, within each block, the people can be arranged in 5!, 5!, and 1! ways. So, yes, multiplying by those factors.

Alternatively, if we think of the problem as first arranging the blocks around the table, which is (3-1)! , then permuting the people within each block. So, the total should be 2 * 5! *5! .
</execution_4.1>

<execution_4.2>
So, the Independent is fixed once the blocks are arranged. Therefore, the total number of arrangements is indeed 2 * 5! * 5! * 1! .

But wait, another way to think about it: if we fix the position of the Independent seat, then the table is effectively linearized, and we can arrange the blocks relative to the Independent. But since it's circular, fixing one position removes the rotational symmetry. Wait, but maybe not. Let me think.

Alternatively, if we consider the table as circular, and the three blocks must be placed around the table. Since the blocks are of different sizes (except the Independent, which is size 1), but actually, the Democrat and Republican blocks are both size 5, so maybe that affects the count. Wait, if two blocks are the same size, does that affect the number of distinct arrangements? Hmm.

Wait, in circular permutations, when arranging objects, if some objects are identical, we adjust the count accordingly. But in this case, the blocks are the ones that are "identical" in terms of being parties, but their sizes are the same (5,5,1). Wait, but the parties themselves are different: Democrat, Republican, Independent. So even though two blocks are size 5, they are different parties, so swapping them would result in a different arrangement. Therefore, the number of arrangements of the blocks is 3! divided by something? Wait, no. Wait, in circular permutations, if all objects are distinct, it's (n-1)! . If some objects are identical, then we divide by the number of identical objects. But here, the blocks are distinct because they are different parties, even if two have the same size. For example, Democrat and Republican are different parties, so even though they have the same number of members, swapping their positions would create a different arrangement. Therefore, the number of ways to arrange the three distinct blocks around the table is (3-1)! = 2! = 2. So that part is correct.
</execution_4.2>

<execution_4.3>
I think the formula is similar to linear arrangements but adjusted for circularity. When arranging blocks in a line, the number of ways is (number of blocks -1)! * permutations within each block. But in a circular arrangement, it's (number of blocks -1)! * permutations within each block. So, yes, that seems to be the case.

So, in this problem, since there are three blocks, the number of ways to arrange the blocks around the table is (3-1)! = 2. Then, within each block, we arrange the members. Since the Democrat block has 5 members, they can be arranged in 5! ways, Republicans in 5! ways, and Independent in 1! way. Therefore, total arrangements: 2 * 5! * 5! * 1! = 2*(5! )^2*1! .

But let me confirm once again. Let's imagine the table as a circle with 11 seats. The three blocks must occupy consecutive seats. Since the table is circular, the blocks can be rotated around the table. However, because the parties are distinct, rotating the entire arrangement would result in a different seating if the blocks are ordered differently. Wait, no. Wait, if we rotate the entire arrangement, the order of the blocks would shift, but since the blocks are of different sizes (except for the Independent, which is size 1), rotating might not preserve the order.

Wait, hold on. Wait, the Democrat and Republican blocks are both size 5, and the Independent is size 1. So, if we have blocks D, R, I arranged in that order, rotating the table by 5 seats would make the D block start where the R block was, but since D and R are the same size, rotating by 5 seats would realign the blocks. Wait, but in that case, the arrangement would be considered the same if it's a rotation. Wait, but the problem states that two seatings are equivalent if one is a rotation of the other. Therefore, even if the blocks are arranged as D-R-I, rotating the table such that the D block moves to where R was would result in the same seating. Therefore, the number of distinct arrangements is not just the permutations of the blocks, but considering that rotating the entire arrangement doesn't count as a different seating.

Wait, this is conflicting with my previous thought. Let me clarify.

If the blocks are considered as a sequence, then arranging them around the table as D-R-I is the same as R-I-D if you rotate the table. But since the blocks are of different sizes (except D and R are both 5), rotating by 5 seats would map D to R's position, but since D and R are the same size, maybe that's allowed. Wait, but the problem says all members of each party sit next to each other, so the blocks must maintain their internal order? Or can they be arranged in any order?

Wait, perhaps I need to model this as arranging the blocks around the table, considering rotational symmetry. So, if we have three blocks, D, R, I, the number of distinct arrangements is (3-1)! = 2 because of rotational symmetry. But if two of the blocks are the same size, does that affect anything? Hmm.

Alternatively, perhaps treating the problem as arranging the three blocks around the table, which is a circular arrangement, so (3-1)! = 2. Then, within each block, arrange the members. So, the total number is 2 * 5! * 5! * 1! .

But let me think of another way. Suppose we "break" the circular table to make it linear. To account for circular permutations, we can fix one position and arrange the rest linearly. So, if we fix the position of the Independent member, then the remaining 10 seats are arranged in a line, with the Democrat and Republican blocks each of 5 seats. But since the table is circular, fixing the Independent member's position effectively makes the arrangement linear with respect to the remaining seats. However, the Democrat and Republican blocks can be arranged in two ways: Democrat first then Republican, or Republican first then Democrat. Then, within each block, the members can be arranged in 5! ways each. So, fixing the Independent's position, we have 2 arrangements for the blocks, and 5! *5! for the members. Therefore, total arrangements: 2*5! *5! .

This approach gives the same result as before. So, by fixing the position of the Independent (since they are single), we eliminate the rotational symmetry. Therefore, the answer is 2*(5! )^2.
</execution_4.3>

<execution_4.4>
5! is 120. So, 120*120=14,400. Then, 14,400*2=28,800. So, 28,800 ways.

But let me think again. Wait, another way to approach this is to consider the circular arrangement of the blocks and then the internal arrangements. So, first, treat each party as a single "super person". Then, arranging three "super people" around a circular table: (3-1)! = 2 ways. Then, within each "super person", arrange the members. For Democrats: 5! , Republicans:5! , Independent:1. So, total arrangements: 2*5! *5! *1! = 2*(5! )^2=28,800.
</execution_4.4>
</Execution_4>

<Plan_5>
Based on execution_4.4:
Is there any chance I missed something? For example, the fact that the Independent is a single seat, does that affect the arrangement? Well, since they are a single person, their position is fixed once the blocks are arranged.

<prompt_5.1> Perform a final verification of the solution and consider any missed factors. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
So, the blocks are D, R, and I. If we fix I's position, then D and R can be arranged in two ways: D next to I on one side and R on the other, or R next to I and D on the other. But since the table is circular, fixing I's position removes the rotational symmetry. Therefore, the two arrangements of D and R are distinct because we can't rotate one into the other. Therefore, that's correct. So, 2 ways for the blocks, then 5! *5! for the internal arrangements. So, yes, 2*(5! )^2.

Alternatively, if we didn't fix the position of the Independent, but instead considered the entire circular arrangement, then the number of ways to arrange the blocks would be (3-1)! = 2, as rotations are considered the same. Then, the internal arrangements are 5! *5! *1! . So, same result.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Therefore, the answer is 2*(5! )^2.

Calculating that:

<prompt_6.1> Calculate the total number of arrangements and format the final answer. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
5! = 120

120^2 = 14,400

14,400 * 2 = 28,800

So, 28,800 ways.
</execution_6.1>
</Execution_6>

<Final_answer>
Therefore, the total number of ways they can sit around a circular table, with all members of each party sitting next to each other, is $\boxed{2 \times (5! )^2}$, which equals $\boxed{28{,}800}$.
</Final_answer>
</think>