TITLE: Let $V=\{1,2,...,n\}$ be a set of vertices. How many directed trees are there over $V$ such that there are exactly 2 vertices with out degree of 0
QUESTION [6 upvotes]: Let V=$\{1,2,...,n\}$ be a set of vertices. How many directed trees are there over V such that there are exactly 2 vertices with out degree of 0?

My first thought was to use symmetry, and that after finding the number of directed trees over $V$ such that $d_{out}(1)=d_{out}(2)=0$, and then multiplying the answer by ${n \choose 2}$.
Using the laplacian matrix seems pointless to me because I have no control over other vertices that will have out degree of $0$.
I know that over a finite number of vertices, it is enough to make sure that there is $r\in V$ such that $d_{in}(r)=0$, and that for all $u\neq r$, $d_{in}(u)=1$, in addition to that that the graph must not have cycles in it.
Now, I tried to use symmetry again. 1 and 2 will never be roots, and by symmetry if I choose a random vertex to be a root, I will need to multiply the result by $n-2$ to get all the possibilities.
However this is pretty tricky as well, the best I got is that this number equals to words over $V$, with a length of $n-1$ with these conditions:

1 and 2 don't appear in the word.
All other vertices appear at least once.
All vertices except the root (and 1 and 2) has a single place in the word where they can't appear.
The root must appear at least once after the first two letters.

Now I am quite stuck however, is there a more simple way to solve this problem that I missed?

REPLY [4 votes]: Here is an easier way to arrive at Marko Riedel's answer.
Since we are dealing with labeled directed graphs, we will need to choose a root. There are $n$ ways to do this. 
Now we must choose two nodes from the remaining $n-1$ nodes that will serve as our leaves. There are $$n-1 \choose 2$$ ways to do this.
Finally we must figure out what to do with the rest of the nodes. Let P denote the path from the root to one of the leaves and Q denote the path from the root to the other leaf. Since there are only two leaves, each of the remaining nodes either is contained in both P and Q, only P, or only Q.
Thus we must count how many ways are there to put $n-3$ nodes into $3$ sets keeping order in mind. Note that this is a classic stars and bars problem with $n-3$ labeled stars and $2$  bars. The number of ways to do this is $$\frac{(n-3+2)!}{2!} = \frac{(n-1)!}{2}$$
Combining these three parts gives us: $$n \cdot {n -1 \choose 2} \cdot \frac{(n-1)!}{2} = \frac{n!(n-1)(n-2)}{4}$$