[blah_blah]

I don't want to spoil it, but in fact, if \sigma is an element of the symmetric group on n letters, p is a polynomial, and \fix\sigma denote the number of fixed points of \sigma, then there are methods to evaluate

\sum_{\sigma\in S_n} p(\fix\sigma)

here is a simple proof of the for the case p = id which generalizes to higher degree polynomials. let \fix_i \sigma = 1 if \sigma fixes the ith place and 0 otherwise.

\sum_{\sigma\in S_n} \fix\sigma =
\sum_{\sigma\in S_n} \sum_i \fix_i\sigma =
\sum_i \sum_{\sigma\in S_n} \fix_i\sigma =
\sum_i (n-1)! =
n(n-1)! =
n!

which is the desired result.