Example: $3 n$ people attend a party, each bringing a hat and an umbrella. After the party, each person randomly takes a hat and an umbrella.
(1) How many possibilities are there such that no one gets back their original items?
(2) How many possibilities are there such that someone gets back their original items?
(3) How many possibilities are there such that exactly 1 person gets back their original items, and the remaining $(n-1)$ people do not get back any of their original items?
(1) $\boldsymbol{n}$ people first take hats, by the problem, none of them take the right one, there are $u_{n}$ ways; then they take umbrellas, by the problem, none of them take the right one again, there are also $u_{n}$ ways. By the multiplication principle, the total number of ways is
$$
u_{n}^{2}=\left[n!\sum_{i=2}^{n} \frac{(-1)^{i}}{i!}\right]^{2} \text { (ways). }
$$
(2) $n$ people, each taking one hat... and one umbrella, the different ways are $(n!)^{2}$ ways, subtracting the case where no one gets back any of their original items gives the desired result. Therefore, the number of ways is
$$
\begin{array}{l}
(n!)^{2}-u_{n}^{2} \\
=(n!)^{2}\left[1-\left(\sum_{i=2}^{n} \frac{(-1)^{i}}{i!}\right)^{2}\right] \text { (ways). }
\end{array}
$$
(3) "Exactly 1 person gets back his original item" this 1 person should be chosen from $n$ people, there are $C_{n}^{1}$ ways to choose. This 1 person taking his original item has 3 possibilities: taking the right hat but the wrong umbrella, taking the wrong hat but the right umbrella, taking both the hat and the umbrella correctly. The first two cases each have $u_{n-1} \cdot u_{n}$ ways, and the third case has $u_{n-1}^{2}$ ways. By the multiplication principle, there are
$$
\begin{array}{l}
C_{n}^{1}\left(2 u_{n-1} u_{n}+u_{n-1}^{2}\right) \\
=n u_{n-1}\left(2 u_{n}+u_{n-1}\right) \\
=n\left[(n-1)!\sum_{i=2}^{n-1} \frac{(-1)^{i}}{i!}\right]\left[2 n!\sum_{i=2}^{n} \frac{(-1)^{i}}{i!}\right. \\
\left.\quad+(n-1)!\sum_{i=2}^{n-1} \frac{(-1)^{i}}{i!}\right] \\
=n!\quad(n-1)!\left[\sum_{i=2}^{n-1} \frac{(-1)^{i}}{i!}\right] \\
\quad \cdot\left[\sum_{i=2}^{n-1} \frac{(2 n+1)(-1)^{i}}{i!}+\frac{2(-1)^{n}}{(n-1)!}\right]
\end{array}
$$
(ways).