Let $n \ge 9$. How many trees are there on vertex set $[n]$ such that at least one vertex has degree $n-3$?

Note: I corrected a huge logical error in my original edit.

It is a little easier to use this generalization of Cayley's formula:

The number of labeled trees on $n$ vertices with degree sequence $(d_1,\dots,d_n)$ is $$\binom{n-2}{d_1-1,d_2-1,\dots,d_n-1}.$$

For trees with one vertex of degree $n-3$, the only possible degree sequences are permutations of one of the following:

• $(n-3,3,1,\dots,1)$.

• $(n-3,2,2,1,\dots,1)$.

To see this, note that for a tree, the sum of $(\deg v-1)$ as $v$ ranges over vertices is $n-2$, and on of the vertices has $\deg v-1=n-4$, so the remaining $(n-2)-(n-4)=2$ can either be given to a single vertex, or two different vertices.

You can then sum up, for each of these sequences, the number of permutations of that sequence, times the corresponding multinomial coefficient. This method has about the same amount of accounting, but not need to draw any actual trees.