<s>[INST] Number of trees $T = (V,E) \;$ such that $V = \left \{ 1,2,3,4,5,6,7 \right \} \;$, $\deg(1) = 2$ , $\deg(t)<5 \;\forall\; 2\leq t\leq 7$ I tried summing all of the cases (I found 4), but somehow i'm getting the wrong answer.
For ex. the number of trees on 7 vertices such that the following sequence is the sequence of their degrees:
My solution (wrong) using Cayley's formula:
The only possible degree sequences are:
$(2,1,1,2,2,2,2) = \binom{5}{1}*\binom{6}{4}$
$(2,3,2,2,1,1,1) = \binom{5}{2}*6\binom{5}{2}$
$(2,3,3,1,1,1,1) = \binom{5}{2,2}*\binom{6}{2}$
$(2,4,2,1,1,1,1) = \binom{5}{3}*6*4$
What am I missing here?
 [/INST] The unlabelled trees with $7$ vertices are
We have to rule out the last three which have a vertex of degree $\geq 5$ or no vertex of degree $2$. 
So for each of the $8$ cases we compute the number of labelled trees such that
 $\deg(1) = 2$ and $\deg(t)<5 \;\forall\; 2\leq t\leq 7$ and we add them  all together:
$$5\cdot \frac{6!}{2}+3\cdot\frac{6!}{2}+3\cdot 6!+3\cdot\frac{6!}{3!}+1\cdot \frac{6!}{2^3}+1\cdot\frac{6!}{2}+2\cdot\frac{ 6!}{3!}+2\cdot\frac{6!}{2^2}=6450.$$
Each term is obtained as the number of vertices of degree $2$ (positions for the vertex $1$) multiplied by $6!$ (for arranging the remaining $6$ vertices) and finally divided by the number of symmetries.
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