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 
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?
 A: 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: $$n \cdot {n -1 \choose 2} \cdot \frac{(n-1)!}{2} = \frac{n!(n-1)(n-2)}{4}$$
A: The combinatorial class of rooted labeled trees with leaves marked is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T} = \mathcal{U} \times \mathcal{Z} +
\mathcal{Z} \times \textsc{SET}_{\ge 1}(\mathcal{T})$$
Which gives the functional equation (EGF)
$$T(z) = uz + z \times (\exp T(z) - 1)$$
so that
$$z = \frac{T(z)}{\exp T(z)-1+u}$$
We then have (this is an OGF in $u$)
$$T_n = (n-1)! [z^{n-1}] T'(z)$$
and from the Cauchy Coefficient Formula
$$(n-1)! [z^{n-1}] T'(z) =
\frac{(n-1)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} T'(z) \; dz.$$
Now put $T(z) = w$ so that $T'(z) \; dz = dw$
and we obtain
$$\frac{(n-1)!}{2\pi i}
\int_{|w|=\gamma}
\frac{(\exp(w)-1+u)^n}{w^n} \; dw.$$
Extracting the count of $k$ leaves where $1\le k\le n-1$
we find
$${n\choose k} \frac{(n-1)!}{2\pi i}
\int_{|w|=\gamma}
\frac{(\exp(w)-1)^{n-k}}{w^n} \; dw
\\ = {n\choose k} \frac{(n-1)!}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^n}
\sum_{q=0}^{n-k} {n-k\choose q} (-1)^{n-k-q} \exp(qw) \; dw
\\ = {n\choose k} (n-1)!
\sum_{q=0}^{n-k} {n-k\choose q} (-1)^{n-k-q}
\frac{q^{n-1}}{(n-1)!} \; dw.$$
We thus have for the answer
$$\bbox[5px,border:2px solid #00A000]{
{n\choose k}
\sum_{q=0}^{n-k} {n-k\choose q} (-1)^{n-k-q} q^{n-1}.}$$
This points us to OEIS A055302 where these
data are confirmed. Observe that this simplifies by inspection to
$$\frac{n!}{k!} \frac{1}{(n-k)!}
\sum_{q=0}^{n-k} {n-k\choose q} (-1)^{n-k-q} q^{n-1}
= \frac{n!}{k!} {n-1\brace n-k}.$$ 
As a sanity  check we should have  obtained all of them.   The sum may
include $k=0$ and $k=n$ because the corresponding Stirling numbers are
zero:
$$\sum_{k=0}^{n} \frac{n!}{k!} {n-1\brace n-k}
= \sum_{k=0}^{n} \frac{n!}{k!}
(n-1)! [z^{n-1}] \frac{(\exp(z)-1)^{n-k}}{(n-k)!}
\\ = (n-1)! [z^{n-1}] \sum_{k=0}^{n} {n\choose k}
(\exp(z)-1)^{n-k}
\\ = (n-1)! [z^{n-1}] \exp(nz) = (n-1)! \frac{n^{n-1}}{(n-1)!}
= n^{n-1}$$
and the check  goes through. We get  for the case of  two leaves the
result
$$\frac{1}{2} n! {n-1\brace n-2} =
\frac{1}{2} n! {n-1\choose 2} $$
or
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{4} n! \times (n-1) \times (n-2).}$$
This will produce starting at three:   
$$3, 36, 360, 3600, 37800, 423360, 5080320, 65318400, \ldots$$
which points to OEIS A055303.
A: Vertices with out-degree = 0 are leaves.
So you have exactly two leaves in a directed tree.
Traverse edges back from those leaves to root (to see how it looks like).
So you order your vertices in a string, then decide who is the root - and you get a tree with two leaves. And leaves can not be roots.
In the end divide by 2, to count for symmetry.
Kind of $\frac{(n-2)\cdot n!}{2}$ different trees.
