Question: Ten persons are seated at random in a row. What is the probability that a particular couple will be seated together?
My attempt: 9! 2!/ 10! = $\dfrac{1}{5}$ , since there are 9! ways of sitting in pairs and 2! ways to arrange a couple.
The solution I'm given is $\dfrac{1}{63}$.
Can someone point out what I'm doing wrong?
Best Answer
You’re doing nothing wrong, assuming that your reasoning is similar to Alex Becker’s in his comment: the correct answer is indeed $\frac15$. Here’s another route to it.
There are $\binom{10}2=45$ pairs of seats, and the couple is equally likely to occupy any one of those $45$ pairs of seats. Nine of the $45$ pairs are adjacent, so the probability that they will occupy adjacent seats is $\frac9{45}=\frac15$, as you say.
And here is yet another. The man sits in an end seat with probability $\frac2{10}=\frac15$. If he’s in an end seat, only one of the remaining nine seats is adjacent to him, and his wife’s probability of getting that seat is $\frac19$. With probability $\frac45$ the man sits in one of the eight seats that have two neighbors, and in that case his wife’s probability of ending up next to him is $\frac29$. The overall probability that the end up sitting together is therefore
$$\frac15\cdot\frac19+\frac45\cdot\frac29=\frac9{45}=\frac15\;.$$
Added: And just for fun, here’s yet another. Imagine that the seats are arranged in a circle around a table. Wherever the wife is sitting, the husband’s probability of sitting next to her is $\frac29$. Then the table is taken away and the seats unwrapped into a straight line, with the breakpoint chosen at random: with probability $\frac1{10}$ it will fall between the husband and the wife, so with probability $\frac9{10}$ they will still end up together. Thus, they end up together with probability $$\frac9{10}\cdot\frac29=\frac15\;.$$