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Problem:

Suppose $n$ couples are seated at a rectangular table with husbands on one side and wives on the other. What is the probability that no husbands are seated directly across from their wives?

Solution attempt:

Denote the husbands by $h_1, \ldots, h_n$ and the wives by $w_{1}, \ldots, w_{n}$. The number of possible arrangements of the couples is $(n!)^2$. To find the number of arrangements in which no husband sits directly across from his wife we can look at one particular arrangement of husbands and multiply the result by $n!$. So fix an arrangement of husbands, say $h_1, \ldots, h_n$. Now, $w_1$ may sit at any of the $n-1$ spots not across from $h_1$. If $w_1$ sits across from $h_2$ then $w_2$ may sit at any of the $n-1$ remaining spots and, if $w_2$ sits across from $h_1$ then we are in the case of $n-2$ couples. Otherwise, if $w_1$ does not sit across from $h_2$ then $w_2$ has $n-2$ possible places to sit. Continue in this way?

I think I'm approaching this incorrectly. This argument leads to too much branching for a problem set this early in the semester. Any help is appreciated!

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1 Answer 1

up vote 1 down vote accepted

Since you can fix the positions of the men, this problems comes down to the question 'what is the probability that a random permutation has no fixed points?'.

The answer to that question can be found here. The answer is obtained using the principle of inclusion and exclusion and comes down to: \begin{align} n!\left(\frac 1{1!}-\frac 1{2!}+\cdots \pm \frac 1{n!}\right)=n!\sum_{k=0}^n\frac{(-1)^k}{k!} \end{align}

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