Derangement counting formula ($!n=n!{\sum }_{k=0}^n\frac{(-1{)}^k}{k!}$)

Inclusion-exclusion applied to the events that a permutation fixes a specified point gives the displayed formula. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ choices of $k$ fixed points and $(n-k)!$ permutations fixing them. The probability that a uniformly chosen permutation is a derangement tends to $e^{-1}$.

// Target: analysis.bigb