I have the identity: $2(n-1)n^{n-2} = \sum_{k=1}^{n-1}{n \choose k}k^{k-1}(n-k)^{n-k-1}$. I am trying to explain this in terms of $T_n$ which is the number of trees on vertex set $[n]$ ($T_n = n^{n-2}$). It looks to me like the right side is counting forests in some context based on other identities I have come across, but I don't have a great understanding of this topic.
Asked
Active
Viewed 62 times
0
-
By way of enrichment, here are two examples of algebraic proofs of these types of identities: MSE link I and MSE link II. – Marko Riedel Sep 17 '15 at 19:12
1 Answers
0
Let $\langle U,V\rangle$ be an ordered partition of $[n]$; for $k=1,\ldots,n-1$ there are $\binom{n}k$ such partitions with $|U|=k$. There are $k^{k-2}$ trees on the vertex set $U$, and for each of them there are $k$ ways to root the tree, so there are $k^{k-1}$ rooted trees on the vertex set $U$. Similarly, there are $(n-k)^{n-k-1}$ rooted trees on the vertex set $V$. Thus, there are $\binom{n}kk^{k-1}(n-k)^{n-k-1}$ ways to split $[n]$ into two non-empty sets and form a rooted tree on the resulting vertex sets. Note, though, that each pair of trees is formed twice, since the splits $\langle U,V\rangle$ and $\langle V,U\rangle$ give rise to the same pairs of trees. Thus, there are altogether
$$\frac12\sum_{k=1}^{n-1}\binom{n}kk^{k-1}(n-k)^{n-k-1}$$
ways to build a pair of non-empty rooted trees on the vertex set $[n]$. Can you see why this must be equal to $(n-1)T_n$? (I’ve put the answer in the spoiler-protected block below.)
Remove an edge from any tree on the vertex set $[n]$; the ends of the removed edges define roots in the two trees that result.
Brian M. Scott
- 616,228