Example 3.19 Find the counting formula for $S_{1}(n, 2)(n \geqslant 2)$.
Solution: Since
$$
\begin{aligned}
S_{1}(n, 2)= & S_{1}(n-1,1)-(n-1) S_{1}(n-1,2) \\
= & (-1)^{n-2}(n-2)!-(n-1) S_{1}(n-1,2) \\
& \quad(n \geqslant 2),
\end{aligned}
$$

Therefore, $(-1)^{n} S_{1}(n, 2)=(n-1)(-1)^{n-1} S_{1}(n-1,2)+(n-2)$!
$$
(n \geqslant 2) \text {, }
$$
$$
\begin{aligned}
\frac{(-1)^{n} S_{1}(n, 2)}{(n-1)!} & =\frac{(-1)^{n-1} S_{1}(n-1,2)}{(n-2)!}+\frac{1}{n-1} \\
& =\frac{(-1)^{n-2} S_{1}(n-2,2)}{(n-3)!}+\frac{1}{n-2}+\frac{1}{n-1} \\
& =\cdots \\
& =\frac{(-1)^{2} S_{1}(2,2)}{1!}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1} \\
& =1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}=\sum_{k=1}^{n-1} \frac{1}{k},
\end{aligned}
$$

That is, $S_{1}(n, 2)=(-1)^{n}(n-1)!\sum_{k=1}^{n-1} \frac{1}{k} \quad(n \geqslant 2)$.