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
that is $\square$
$$\begin{array}{l}
C_{s}^{r-1}+C_{s}^{r}=C_{s+1}^{r} \\
C_{s}^{r}=C_{s+1}^{r}-C_{s}^{r-1}
\end{array}$$

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 the equations, we get
$$\begin{array}{ll} 
& C_{m 0}^{0}+C_{m+1}^{1}+C_{m+2}^{2}+\cdots+C_{m+n}^{n}=C_{m+n+1}^{n}, \\
\text { Therefore } \quad S=m!C_{m+n+1}^{n}=\frac{(m+n+1)!}{(m+1) \cdot n!} \text {. }
\end{array}$$