Proof concerning Stirling numbers of the first kind How do I show that $${ n\brack 2} = \frac{n!}{2} \sum_{m=1}^{n-1} \frac{1}{m(n-m)}$$ using a combinatorial argument? My definition of ${n \brack 2}$ is " _the number of elements of $S_n$ which can be decomposed as the product of exactly $2$ disjoint cycles_ ".

The (standard) combinatorial interpreation of $\begin{bmatrix}n\\\ i\end{bmatrix}$ is as permutations of $n$ into $i$ (disjoint) cycles; we are interested in the case $i=2$.

Consider a permutation of $n$ into two cycles of length $\color{red}{m}$ and $\color{blue}{n-m}$
\begin{eqnarray*} \color{red}{(a_1 a_2 \cdots a_m)} \color{blue}{(b_1 b_2 \cdots b_{n-m})}. \end{eqnarray*} The set $n$ can be partitioned into two (disjoint) sets of size $\color{red}{m}$ and $\color{blue}{n-m}$ in $\binom{n}{m}$ ways, the elements in the first cycle can be arranged in $\color{red}{(m-1)!}$ ways & the elements in the second cycle can be arranged in $\color{blue}{(n-m-1)!}$ ways, as $m$ ranges from $1$ to $n-1$ we will double count the possible configurations, thus \begin{eqnarray*} \begin{bmatrix}n\\\ 2\end{bmatrix} = \frac{1}{2} \sum_{m=1}^{n-1} \binom{n}{m} \color{red}{(m-1)!} \color{blue}{(n-m-1)!} = \frac{n!}{2} \sum_{m=1}^{n-1} \frac{1}{\color{red}{m}\color{blue}{(n-m)}}. \end{eqnarray*}