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.  

This seems to be explained in this wikipedia article.
