Number of involution in symmetric group Can any one show me on how to prove that the number of involutions in the symmetric group on $n$ letters, $$\sum_{k=0}^{\lfloor n/2 \rfloor} {n \choose 2k} (2k-1)!! = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{2^kk!(n-2k)!}$$
I've tried proving it via Young tableaux, knowing that the number of tableaux is the number of involutions, but I haven't got the right answer so far.  

Let $\tau$ be an involution. Writing $\tau$ as a product of disjoint cycles, we see that no cycle can have length greater than $3$ since $\tau$ has order $2$. It follows that $\tau$ is a product of disjoint transpositions. 
Suppose that $\tau$ is a product of $k$ disjoint transpositions. Then it permutes precisely $2k$ letters so we must choose those letters out of the $n$ available. Next we must group the $2k$ letters into $k$ pairs. There are precisely
$$a_k = \binom{n}{2k}\frac{(2k)!}{k!2^k} = \binom{n}{2k}(2k-1)!!$$
ways to do this so there are precisely $a_k$ involutions with $k$ disjoint transpositions. 
Clearly $k$ can range from $0$ (the identity) to $\lfloor \frac{n}{2}\rfloor$ transpositions. Summing through the different values of $a_k$ gives the desired result.
