Number of undirected trees under conditions

Question

I'm trying to calculate the number of undirected trees with $8$ labeled vertices for each of the conditions:

1. vertices $2$ and $6$ have the same degree.

2. There are exactly $3$ leaves.

Answer 1

Solution for 2). A tree with $8$ vertices has $7$ edges. Therefore, if it has exactly 3 leaves then $$\sum_{d\geq 2}dn_d+3 = \sum_{v\in V}\text{deg}(v)=2\cdot 7=14$$ where $n_d$ is the number of vertices of degree $d$ with $\sum_{d\geq 2}n_d=8-3=5$. It follows that the unique solution is: $n_2=4$, $n_3=1$ and $n_i=0$ for $i>3$. So the tree has one vertex of degree $3$ with three branches of lengths of one of these types: $(5,1,1)$, $(4,2,1)$, $(3,3,1)$, $(3,2,2)$.

Now we count the labelings: $$(5,1,1)\to \frac{8!}{2}\;,\quad (4,2,1)\to 8!\;,\quad (3,3,1)\to \frac{8!}{2}\;,\quad (3,2,2)\to \frac{8!}{2}.$$ Therefore the total number is $5/2\cdot 8!=100800.$

P.S. The 23 trees with 8 unlabeled vertices can be seen here.