Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 2 / 9F / Solution

The inclusion-exclusion principle givesFor the coat problem, all $n!$ permutations are equally likely. If a specified set of $j$ people gets its own coats, the other coats can be assigned in $(n-j)!$ ways, so that intersection has probability $(n-j)!/n!$. Taking the complement of the union of correct-coat events gives the derangement probabilityTo have exactly $m$ fixed coats, choose their recipients and derange all remaining coats. Thus the fixed point count of a uniform random permutation hasand zero probability outside this range. The empty permutation has $D_0=1$, so the formula includes $m=n$; it also gives zero for $m=n-1$, since a lone remaining coat cannot be misplaced.

For fixed $m$, $n-m\to \infty$, and the exponential series givesThese are the probabilities of a Poisson distribution of mean one. The limit is for fixed $m$, not for $m$ increasing with $n$.