How many trees are there on the vertex set $\lbrace{1,2,\ldots,7\rbrace}$ in which the vertices $2$ and $3$ have degree 3, and vertex $5$ has degree 2, and all the others have degree 1.
I proceeded as follows:
We know that vertices $2$ and $3$ each has to appear twice in Prüfer code, and vertex $5$ has to appear once. Since the Prüfer code has length $n-2$, where $n =7$, our answer should be the same as counting the numbers of ways of ordering $5$ objects, where each object falls into 1 of 3 indistinguishable categories of sizes 2,2, and 1. Then this is the same as the multinomial coefficient
\begin{equation} {5\choose{2,2,1}}=\dfrac{5!}{2!2!1!} = 30. \end{equation}
I think my answer is correct but I was wondering if there's another way of solving this problem without using the Prüfer code?
For the moment, disregard the vertices of degree 1. So $\deg v_3=3$, $\deg v_3=3$, and $\deg v_5=2$. How many ways could these be put together? Because they all must be in the same connected component (it's a tree), we know that they must all be adjacent. You could connect them in one of 3 distinct orders:
Now we attach the vertices of degree 1.
In the first case, of the remaining vertices, we must choose 2 to attach to $v_2$, 1 to attach to $v_3$, and then the remaining 1 to attach to $v_5$. There are $\binom{4}{2}\binom{2}{1}\binom{1}{1}=12$ ways to do this.
Similarly, in the second case, we choose two to attach to each of $v_2$ and $v_3$, and none to attach to $v_5$, yielding 6 options.
In the third case, we choose two to attach to $v_3$, one to attach to $v_2$, and 1 to attach to $v_5$, yielding 12 as in the first case.
All together, this gives $12+12+6=30$ options.