= Solution
The <inclusion-exclusion principle> gives
$$
\boxed{\mathbb P\left(\bigcup_{i=1}^nA_i\right)=
\sum_{\varnothing\ne I\subseteq\{1,\ldots,n\}}(-1)^{|I|+1}\mathbb P\left(\bigcap_{i\in I}A_i\right).}
$$
For 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> <probability>
$$
\boxed{p(0,n)=\sum_{j=0}^n\frac{(-1)^j}{j!}.}
$$
To have exactly $m$ fixed coats, choose their recipients and derange all remaining coats. Thus the <fixed point count of a uniform random permutation> has
$$
\boxed{p(m,n)=\frac{\binom nmD_{n-m}}{n!}
=\frac{p(0,n-m)}{m!}
=\frac1{m!}\sum_{j=0}^{n-m}\frac{(-1)^j}{j!},\quad0\le m\le n,}
$$
and 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 gives
$$
\boxed{\lim_{n\to\infty}p(m,n)=\frac{e^{-1}}{m!}.}
$$
These are the <probabilities> of a <Poisson distribution> of <mean> one. The limit is for fixed $m$, not for $m$ increasing with $n$.
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