3・12 Try to find the sum:
$$
S=m!+\frac{(m+1)!}{1!}+\frac{(m+2)!}{2!}+\cdots+\frac{(m+n)!}{n!} .
$$
[Solution] $S=m!+\frac{(m+1)!}{1!}+\frac{(m+2)!}{2!}+\cdots+\frac{(m+n)!}{n!}$
$$
\begin{aligned}
= & m!\left[1+\frac{m+1}{1!}+\frac{(m+2)(m+1)}{2!}+\cdots+\right. \\
& \left.+\frac{(m+n)(m+n-1) \cdots(m+2)(m+1)}{n!}\right] \\
= & m!\left(C_{m}^{0}+C_{m+1}^{1}+C_{m+2}^{2}+\cdots+C_{m+n}^{n}\right),
\end{aligned}
$$

Using the combinatorial identity $C_{s}^{r-1}+C_{s}^{r}=C_{s+1}^{r}$,
that is,
$$
C_{s}^{r}=C_{s+1}^{r}-C_{s}^{r-1} .
$$

Thus, we have
$$
\begin{array}{l}
C_{m}^{0}=C_{m+1}^{0}, \\
C_{m+1}^{1}=C_{m+2}^{1}-C_{m+1}^{0}, \\
C_{m+2}^{2}=C_{m+3}^{2}-C_{m+2}^{1}, \\
\cdots \cdots \\
C_{m+n}^{n}=C_{m+n+1}^{n}-C_{m+n}^{n-1},
\end{array}
$$

Adding all these equations, we get
$$
C_{m 0}^{0}+C_{m+1}^{1}+C_{m+2}^{2}+\cdots+C_{m+n}^{n}=C_{m+n+1}^{n},
$$

Therefore, $\quad S=m!C_{m+n+1}^{n}=\frac{(m+n+1)!}{(m+1) \cdot n!}$.