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Here's my question and possible answer.

How many possible ways can you arrange 8 married couples between 2 circular tables of 8 identical chairs each such that:

1) each couple must sit at the same table, and,

2) at each table, men and women must sit in adjacent chairs (NOTE: a couple can sit next to each other but doesn't have to).

My solution:

Number of ways = (number of ways to split 8 couples into 2 tables of 4 couples each) * (number of arrangements at each table)

$=\frac{8!}{4!4!}*$(4 men and 4 women sitting alternately in 2 ways)

$=\frac{8!}{4!4!}\!\cdot\! 4!\!\cdot\!4!\cdot\!2$

$=2 * 8!$

I feel that I'm wrong about this.

Can someone verify this solution or provide the correct one?

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First, pick $4$ couples out of the $8$ couples to sit at one table: $8 \choose 4$

Note that this will fix the people at the other table as well. Now, if we differentiate between the two tables, then the number of ways to split the $16$ people between the two tables is ${8 \choose 4} $ (that is the number of ways to pick the people for table 1, fixing the rest for table 2). If you do not differentiate between the tables, then divide this by $2$.

Now let's arrange the people. We'll calculate the number of ways to seat the people around one table, and just multiply by that number again for the other table at the end.

Since it's a circular table with identical chairs, we'll 'anchor' the seats with $1$ of the women. Then, we can seat the other women in $3!$ ways relative to this woman, and the men in $4!$ ways.

Total: $${8 \choose 4} \cdot 3! \cdot 4! \cdot 3! \cdot 4!$$

And again, if you do not differentiate between the tables, then divide this by $2$

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    $\begingroup$ Suppose we have an origin from where to start counting round one of the tables, then this would class the string ABCDEFGH to be different to BCDEFGHA, but surely due to the question detailing that the tables are circular, these are the same? Everyone is sitting next to the same people either side. Also, you may split up a couple if you make the selections by choosing women first, then men. It would be better to pair up the couples and choose 4 couples to sit at a table. $\endgroup$ – John Doe Oct 29 '17 at 0:10
  • $\begingroup$ @JohnDoe Right, that's why you get $3! \cdot 4!$ ways to seat the people around the table, instead of $4! \cdot 4!$. I do like your idea of first pairing up the men and women, and then seating them one pair at a time: you should create your own answer using that method! But note: you will get that same $3!$ factor when seating the 4 pairs around the circular table $\endgroup$ – Bram28 Oct 29 '17 at 0:16
  • $\begingroup$ My answer disagreed with yours by a factor of $8\choose4$, due to how I chose the couples. I am fairly sure mine is correct, but feel free to enlighten me if I am missing something. $\endgroup$ – John Doe Oct 29 '17 at 0:27
  • $\begingroup$ @JohnDoe I didn't check tour answer, but my answer is certainly wrong! I forgot to account for the fact that a couple needs to sit at the same table! $\endgroup$ – Bram28 Oct 29 '17 at 0:29
  • $\begingroup$ Ah, yes that's why mine is different! Ours should agree once you correct this haha $\endgroup$ – John Doe Oct 29 '17 at 0:31
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How I would do it:

How many ways can you split the 8 couples into two groups of 4? That's ${8\choose4}=70$.

Then once split, you have 4 men, 4 women who need to be seated at one of the tables, in alternating order. The first person seated has 8 choices, then the other people of the same gender have 3,2,1 choices afterwards. Meanwhile for the opposite gender, they have 4,3,2,1 choices for their picks. So in total, $8\times4\times3\times3\times2\times2=1152$. Then to account for the rotational symmetry, we must divide by $8$ to give $144$ ways to arrange these couples on their table.

Then it is the same for the other table, so another $144$ ways to arrange them, hence $144^2$ ways to arrange all the people once tables have been chosen for each couple.

So the overall answer is $$70\times144^2=1,451,520$$

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  • $\begingroup$ Nice, I agree with the solution! However it's worth noting that if you fix a person to a particular place you avoid the multiplying and dividing by 8, don't you think? $\endgroup$ – Thomas Bladt Oct 29 '17 at 0:48
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    $\begingroup$ Yeah, multiplying and dividing by 8 was a bit pointless in this question. I suppose it could be useful to see it done like this if an example ever came up where it was required to seat in a line rather than in a circle - then you would still have to multiply by 8, but since there is no rotational symmetry, you wouldn't divide by it. $\endgroup$ – John Doe Oct 29 '17 at 0:50

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