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Five married couples are randomly paired at a rectangular table. Five wives on one side and five husbands on the other side. What is the probability that only one man soy across from his wife.

Here's what I got. I assume A-1 B-2 C-3 D-4 and E-5 are couples. Where letters are wives and numbers are husband.

Suppose a-1 sit across. Then b only has 3 choices because she can't sit across her husband. So as c, who only has 3 choices. Then d has 2 choices and e had the pair up with whatever's left. So the probavility I got is (5* 3*3*2*1 ) / 5!

Am I right?

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Your argument is not correct. When they make choices in succession the number of options of each person depends on the choices made before. E.g., if B chooses 4 then C only can choose among 2 and 5.

We have to count the number of permutations $\pi:\>[5]\to[5]$ with exactly one fixed point. There are $5$ ways to choose the fixed point $f$. Given $f$ we have to count the permutations of the four-element set $[5]\setminus\{f\}$ without fixed points. Such permutations are called derangements; there are $9$ of them. As there are $120$ permutations in all the requested probability comes to $$p={5\cdot 9\over 120}={3\over8}\ .$$

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Yes you're totally right! very good work. You could even simplify it to 3/4, if you would write it out. The exercise would be even easier if you would use Burnside's Lemma.

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