TITLE: Number of trees with a fixed edge
QUESTION [5 upvotes]: Consider a vertex set $[n]$. By Cayley's theorem there there are $n^{(n-2)}$ trees on $[n]$, but how can one count the following slightly modified version: 
What is the number of trees on $[n]$ vertices where the edge $\{1,2\}$ is definitely contained in the trees?

REPLY [8 votes]: We can actually do it directly from Cauchy’s theorem, without making use of a proof of that result.
For each $e=\{k,\ell\}\in[n]$ with $k\ne\ell$ let $S_n(e)$ be the number of trees on $[n]$ that contain the edge $e$, let $e_0=\{1,2\}$, and let $S_n=S_n(e_0)$; clearly $S_n(e)=S_n$ for all $e\in E$, where $E$ is the set of possible edges. Then
$$\sum_{e\in E}S_n(e)=\binom{n}2S_n$$
counts each tree on $[n]$ $n-1$ times, once for each of its edges. There are $n^{n-2}$ trees on $[n]$, so
$$\binom{n}2S_n=(n-1)n^{n-2}\;,$$
and
$$S_n=2n^{n-3}\;.$$

REPLY [2 votes]: Lemma 6 in this paper gives a stronger result: the number of labeled spanning trees on $n$ vertices containing any fixed forest. In your case of a single edge forest, the formula simplifies to $2n^{n-3}$.