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PROBLEM G4 In a couples therapy session, n couples are to be seated at a round table (in 2n chairs), but no person is allowed to sit next to his/her spouse. How many seat assignments are there? Here's the solution: Pick $k$ out of $n$ couples. Then the number of seatings so that these $k$ couples sit together is computed as follows: there are $2n$ ways to seat the first couple (up to order), and then the number of seatings of the rest of the $k$ couples equals $(2n-k-1)...(2n-2k+1)$. So the total equals: $N_k = 2^k \times 2n \times (2n-k-1)!$ Thus, using the inclusion and exclusion formula, we get that the answer is: $N = \sum_{k \ge 0} (-1)^k {n \choose k} N_k$

Here is what I do not understand, if these $k$ couples sit together, then how come is "the number of seatings of the rest of the $k$ couples equals $(2n-k-1)...(2n-2k+1)$", isn't that choosing seats out of all $2n$ seats left, instead of $k$ seats together as a whole block? And even it really is $(2n-k-1)...(2n-2k+1)$, shouldn't it be $A_{2n-2k+1}^{2n-k-1}$? How did it change into $(2n-k-1)!$ in the latter formula?

user330984
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    I think when they mean "these $k$ couples sit together" they mean within each of the $k$ couples, the spouses sit side by side. But the $k$ couples might not be next to one another in the circle. As for why $(2n-k-1)\cdots (2n-2k+1)$ becomes $(2n-k-1)!$, I suspect that has to do with the fact that there are $(2n-2k)!$ ways to seat the people not among the $k$ couples chosen. I admit that I don't completely follow the answer though. – kccu Jun 17 '16 at 12:40
  • See also http://math.stackexchange.com/questions/1815975. – joriki Jun 17 '16 at 12:59

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kccu has basically answered the question in a comment. Here "these $k$ couples sit together" means that the spouses of each couple sit next to each other, not that all $k$ couples sit contiguously. And the missing factor $(2n-2k)!$ is indeed the number of ways of permuting the members of the remaining $n-k$ couples. It would have been simpler to just say that the $2n-k-1$ entities ($k-1$ couples and $2n-2k$ members of the remaining couples) can be permuted in $(2n-k-1)!$ ways.

joriki
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