Nobody gets their own hat

Idea. “Nobody matches” is the complement of a union of events, and the intersections of those events are easy to count. Inclusion–exclusion does the rest.

Step 1: intersections. For any set of $k$ people, the probability that all of them get their own hats is $\frac{(5-k)!}{5!}$. There are $\left(\genfrac{}{}{0ex}{}{5}{k}\right)$ such sets, so the $k$-th term of inclusion–exclusion is $$\left(\genfrac{}{}{0ex}{}{5}{k}\right)\frac{(5-k)!}{5!}=\frac{1}{k!}.$$

Step 2: inclusion–exclusion. $$\begin{array}{rl}\mathbb{P}(\text{no match})& =\sum _{k=0}^5\frac{(-1{)}^k}{k!}\\ & =1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}\\ & =\frac{44}{120}=\frac{11}{30}.\end{array}$$

So $44$ of the $120$ assignments give nobody their own hat, about $36.7\%$.

In general. With $n$ people the probability is ${\sum }_{k=0}^n\frac{(-1{)}^k}{k!}$, which converges to $e^{-1}\approx 0.368$ very fast: from $n=5$ on it is already within $0.002$. The expected number of people who get their own hat is exactly $1$ for every $n$, by linearity.