We define a simple graph $G = \langle V, E\rangle$ to be a tree iff each $v \in V$ is reachable from every other $u \in V$ and also $G$ doesn't include any simple circles.
How many trees $T = \langle V, E \rangle$ exist such that $V =$ { $v_1, v_2, \dots, v_7$} and $deg(v_1) = 1$, $deg(v_j) \leq 5$ for every $j = 2, .., 7$ ?
This is the first time I encounter this type of question, however, the combinatoric underlying problem seems to involve double factorial: We can divide $v_1, \dots, v_6$ into different pairs, .e.g {$v_1, v_2$}, {$v_3, v_4$}, {$v_5, v_6$} and then do the same, considering that each edge (except $v_1$) can be connected to the other from both sides.
However, in the way I proposed it seems that there are different cases to consider (.e.g what are the options for $v_7$), and additionally I struggle to see how I can ensure that I don't count the same options several times.
I would like to get some ideas\advices about it. Thanks
Two hints:
For each $j\in \{2,\dots,7\}$, determine the number of trees on $\{1,\dots,7\}$ such that $\deg v_j\ge 6$. Verify that these trees all have $\deg v_1=1$, as well.
Notice that the number of trees on the vertex set $\{1,2,\dots,7\}$ for which $\deg v_1=1$ is equal to the number of rooted trees on the vertex set $\{2,\dots,7\}$. There is a simple bijection between these two sets.
Using these two points, you should be able to reduce your counting problem for trees on $\{1,\dots,7\}$ to a simpler problem about trees on $\{2,\dots,7\}$, minus a small number of exceptional cases.