$$
\begin{array}{l}
\text { Four, (20 points) Let } x_{2}=1, x_{3}=2, \\
x_{n}=(n-1)\left(x_{n-1}+x_{n-2}\right)(n \geqslant 4) .
\end{array}
$$

Find the general term formula for $\left\{x_{n}\right\}$.
Let's make the substitution $x_{k}=k!y_{k}$, and substitute it into the given recurrence relation to get $k y_{k}=(k-1) y_{k-1}+y_{k-2}$, that is,
$$
y_{k}-y_{k-1}=-\frac{1}{k}\left(y_{k-1}-y_{k-2}\right) \text {. }
$$

Taking $k=4,5, \cdots, m$ in equation (1) and multiplying the corresponding equations.
Noting that $y_{3}=\frac{1}{3}, y_{2}=\frac{1}{2}$, we have
$$
y_{m}-y_{m-1}=(-1)^{m} \frac{1}{m!} \text {. }
$$

Taking $m=4,5, \cdots, n$ in the above equation and adding the corresponding equations, we get
$$
\begin{array}{l}
y_{n}-y_{3}=\sum_{m=4}^{n} \frac{(-1)^{m}}{m!} \\
\Rightarrow y_{n}=\sum_{m=2}^{n} \frac{(-1)^{m}}{m!} \\
\Rightarrow x_{n}=n!\sum_{m=2}^{n} \frac{(-1)^{m}}{m!} .
\end{array}
$$