# Probability married couple problem

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?

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 $$p={5\cdot 9\over 120}={3\over8}\ .$$