Recurrence relation for pairing off $2n$ people I know the answer is supposed to be 
$$a_{2n} = (2n-1) a_{2n-2}$$
Can someone please explain why shouldn't be having $\binom{2n}{2}$ in place of $2n-1$?
Doesn't it matter which two people are paired off out of the $2n$ people and hence generating a different case each time for the remaining $(2n-2)$ people?

Consider the problem of pairing $2n$ people AND label one of these pairs.  Then the number of ways is
$$b_{2n}=\binom{2n}{2}a_{2n-2}$$
This is what we enumerate when in your formula we have $\binom{2n}{2}$ in place of $2n-1$.
Since there are $n$ possible pairs, we may "unlabel" the labelled pair by dividing by $n$:
$$a_{2n}=\frac{b_{2n}}{n}=\frac{1}{n}\binom{2n}{2}a_{2n-2}=(2n-1)a_{2n-2}.$$
