Example 16 Try to express $\sum_{k=0}^{n} \frac{(-1)^{k} \mathrm{C}_{n}^{k}}{k^{3}+9 k^{2}+26 k+24}$ in the form $\frac{P(n)}{Q(n)}$. Here $P(n) 、 Q(n)$ are two polynomials with integer coefficients.
Let the original expression be $S(n)$, then
$$
\begin{array}{l}
S(n)=\sum_{k=0}^{n} \frac{(-1)^{k} \mathrm{C}_{n}^{k}}{k^{3}+9 k^{2}+26 k+24} \\
=\sum_{k=0}^{n} \frac{(-1)^{k} n!}{k!(n-k)!(k+2)(k+3)(k+4)} \\
=\sum_{k=0}^{n} \frac{(-1)^{k}(n+4)!}{(k+4)!(n-k)!} \text {. } \\
{\left[\frac{k+1}{(n+1)(n+2)(n+3)(n+4)}\right] \text {. }} \\
\text { Let } T(n)=(n+1)(n+2)(n+3)(n+4) S(n) \\
=\sum_{k=0}^{n}\left[(-1)^{k+4} \mathrm{C}_{n+4}^{k+4}(k+1)\right] \\
=\sum_{k=-4}^{n}(-1)^{k+4} \mathrm{C}_{n+4}^{k+4}(k+1)- \\
{\left[-3+2(n+4)-C_{n+4}^{2}\right] \text {. }} \\
\end{array}
$$

Let $j=k+4$, then
$$
\begin{array}{l}
T(n)=\sum_{j=0}^{n+4}(-1)^{j} C_{n+4}^{j}(j-3)+ \\
\quad \frac{1}{2}\left(n^{2}+3 n+2\right) \\
=\sum_{j=0}^{n+4}(-1)^{j} j C_{n+4}^{j}-3 \sum_{j=0}^{n+4}(-1)^{j} \mathrm{C}_{n+4}^{j}+ \\
\frac{1}{2}\left(n^{2}+3 n+2\right) .
\end{array}
$$

By the identities (III) and (VI), we have
$$
\begin{array}{l}
\sum_{j=0}^{n+4}(-1)^{j} \mathrm{C}_{n+4}^{j} \\
=\sum_{j=1}^{n+4}(-1)^{j}(n+4) \mathrm{C}_{n+3}^{j-1}=0, \\
\sum_{j=0}^{n+4}(-1)^{j} \mathrm{C}_{n+4}^{j}=0 .
\end{array}
$$

Therefore, $T(n)=\frac{1}{2}\left(n^{2}+3 n+2\right)$.
Then $S(n)=\frac{T(n)}{(n+1)(n+2)(n+3)(n+4)}$
$$
=\frac{1}{2(n+3)(n+4)} \text {. }
$$

Hence $\frac{P(n)}{Q(n)}=\frac{1}{2(n+3)(n+4)}$ is the desired result.