Regarding the following question's answer : Seeking a combinatorial proof of $2n^{n-3} = \sum_{m=1}^{n-1}\binom{n-2}{m-1}m^{m-2}(n-m)^{n-m-2}$
I have some additional questions:
1. why m can't be 0 or n?
2. Plus what (n-2 / m-1) refers to? why are we choosing m-1 from n-2?
Consider trees on the vertex set $[n]=\{1,2,\ldots,n\}$. Some of these trees contain the edge $\{1,2\}$. Let $T$ be such a tree. Let $V_1$ be the set of vertices that can be reached from vertex $1$ without going through $2$, and let $V_2$ be the set of vertices that can be reached from vertex $2$ without going through $1$; note that $1\in V_1$ and $2\in V_2$, that $|V_1|+|V_2|=n$, and that the subgraphs of $T$ induced by $V_1$ and $V_2$ are both trees.
Let $m=|V_1|$. Then $m\ge 1$, since $1\in V_1$, and $m\le n-1$, since $2\notin V_1$. For a specific value of $m\in\{1,2,\ldots,n-1\}$, how many of these trees are there with $|V_1|=m$? We know that $1\in V_1$, so there must be $m-1$ other members of $V_1$. Moreover, $2\notin V_1$, so these $m-1$ members of $V_1$ must be chosen from the $n-2$ vertices $3,4,\ldots,n$. Thus, there are $\binom{n-1}{m-1}$ ways to choose the remaining $m-1$ vertices in $V_1$. There are then $m^{m-2}$ different trees that can be made with those vertices. Finally, there are $n-m$ vertices in $V_2$, so there are $(n-m)^{n-m-2}$ different trees that can be made with those vertices. Putting the pieces together, we see that there are
$$\binom{n-2}{m-1}m^{m-2}(n-m)^{n-m-2}$$
trees on the vertex set $[n]$ that contain the edge $\{1,2\}$ and have $m$ vertices in $V_1$. Summing over the possible values of $m$, we find altogether
$$\sum_{m=1}^{n-1}\binom{n-2}{m-1}m^{m-2}(n-m)^{n-m-2}$$
trees on the vertex set $[n]$ that include the edge $\{1,2\}$.
Alternatively, we can start with the fact that there are $n^{n-2}$ trees on the vertex set $[n]$. Each of them has $n-1$ edges, so there are altogether $(n-1)n^{n-2}$ edges in these trees. We now count these edges in a different way.
Let $t$ be the number of trees containing the edge $\{1,2\}$; then there are $t$ trees containing any given edge, and there are altogether $\frac{n(n-1)}2$ different possible edges, so the total number of edges in all of these trees must be $\frac{n(n-1)}2\cdot t$. Thus,
$$\frac{n(n-1)}2\cdot t=(n-1)n^{n-2}\;,$$
and therefore $t=2n^{n-3}$. And $t$ is the quantity that we computed in the first part of the answer, so we conclude that
$$2n^{n-3}=\sum_{m=1}^{n-1}\binom{n-2}{m-1}m^{m-2}(n-m)^{n-m-2}\;,$$
as desired.