How many labeled trees are there if there're $4$ leaves that are set? 
Let $T=(V,E)$ be a tree of $10$ labeled nodes numbered $1,2,3,\dots,10$ such that there're $4$ labeled leaves $1,2,3,4$ (there may be more leaves in addition to $1,2,3,4$). How many such trees are there and how many trees are there where the leaves are only $1,2,3,4$?

I'll start with the second part. If there're exactly $4$ leaves then for every node except for nodes $1,2,3,4$ we have: 
$$2\le \deg(v)\le 4$$
because if $\deg(v)=1$ then it's an additional leave and if $\deg(v)>4$ then there're more than $4$ leaves.
According to the handshaking lemma:
$$
\sum_{v\in V}\deg(v)=2|E|=2(|V|-1)=18
$$
We now need to find all possible combinations with repetitions of integers $2,3,4$ over $6$ bins ($6$ nodes remain after removal of $4$ leaves) such that their sum is $18-4=14$:
$$
x_1+x_2+x_3+x_4+x_5+x_6=14,\quad 2\le x_i\le 4
$$
which can be solved using generating functions:
$$
(x^2+x^3+x^4)^6=14\\x^{12}(1+x+x^2)=14
$$
so we need to find the coefficient of $x^{2}$ in:
$$
\bigg(\frac{1-x^3}{1-x}\bigg)^6=\sum_{k=0}^6{6\choose k}(-1)^kx^{3k}\cdot \sum_{i=0}^{\infty}{i+5\choose i}x^6
$$
but it's impossible to find the coefficient from this because the powers of $x$ are too high.
I would've multiplied the result by $6!$ to account for order but I'm obviously doing something wrong. 
And regarding the first part I would've found the number of labeled trees with at least the leaves $1,2,3,4$ by first choosing some ${6\choose 3}$ from $6$ nodes that are not $1,2,3,4$ and then using the same approach as above to count all combinations with repetitions and then subtract all that from the total $n^{n-2}$ possible labeled trees.
 A: For a labeled tree on $n$ vertices  the count of those where the first
$m$      are      leaves      is     obtained      using      Pruefer
codes, which have
the property  that the degree of  a vertex in the  tree resulting from
the code is one more than the  number of times it appears in the code.
Thus the fact that  the $m$ nodes are leaves means  they do not appear
in the Pruefer code. Hence there remain
$$(n-m)^{n-2}$$
possible codes.  If these $m$ leaves  are the only ones  all the other
nodes must appear in the Pruefer code. We get using Stirling numbers
$$(n-m)! \times {n-2\brace n-m}.$$
Remark. I noticed the link to the OEIS only now where OP says they
are not familiar  with Stirling numbers.  I trust  the Wikipedia entry
on Pruefer codes is sufficient. With the Stirling numbers we partition
the $n-2$  slots in the Pruefer  code into sets for  the $n-m$ values,
which correspond  to the nodes  and are  then guaranteed to  appear at
least once.  Here we have ordered set partitions however and hence the
multiplier of $(n-m)!.$  E.g. when we fill four slots  with two values
$1$ and  $2$ and have a  set partition into two  sets namely $\{1,3\}$
and $\{2,4\}$  then placing $1$  in the former  and $2$ in  the latter
slots is not the same as $2$ in the former and $1$ in the latter.  The
fact that  Stirling numbers count  unordered set partitions  into into
non-empty sets is all we need to know here.
 Addendum.  OP asks for  clarification of closed form  at OEIS
A055316.  Explanation  is simply  that OEIS
lists the formula  for the number of trees having  exactly $m$ leaves,
as opposed to trees where nodes $1$ to $m$ are the set of leaves. This
means  we need  to choose  the $m$  leaves from  the $n$  nodes first,
getting
$${n\choose m} \times (n-m)! \times {n-2\brace n-m}
= \frac{n!}{m!} \times {n-2\brace n-m}.$$
A: The generating functions are overkill for finding out how 6 numbers which are all 2, 3, or 4 can add up to 14. At minimum, six such numbers sum to 12, so you have 2 excess: either one 4 and five 2s, or two 3s and four 2s.
In the first case, you have four arms from a single vertex of degree 4, and you can vary the lengths of the arms for a total of 10 vertices. In the second case, you have two vertices of degree three connected by a path with zero to four vertices of degree 2, and the remaining two neighbors of each of the degree-3 vertices begin four arms to the leaves.
As far as the generating functions go, there seems to be a missing power of 6 in this line:
$$x^{12}(1+x+x^2)=14$$
but I think that was corrected in the following lines.
