Probability: People sitting in a row (linear arrangement) 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?
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 $$\frac9{10}\cdot\frac29=\frac15\;.$$
There are $10!$ ways how to put these $10$ persons in the row, the couple can seat this way 
$12$, $23$, $34$, $45$, $56$, $67$, $78$, $89$, $10 \ 1$ or $21$, $32$, $43$, $54$, $65$, $76$, $87$, $98$, $1\ 10$ so $20$ possibilities but there are $8$ other persons, there are $8!$ possibilities to seat them so the probability is

$$P = 20 \cdot \frac{8!}{10!} = \frac{2}{9} = 22.2\%$$