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1) How many trees on these vertex have at least 2 leaves ?

2) How many trees on these vertex have more than 2 leaves ?

Attempt -

I will use inclusion-exclusion for both cases -

$1)$ $$\binom{8}{2}\cdot6^6-\binom{8}{3}\cdot5^6 + \binom{8}{4}\cdot4^6 - \binom{8}{5}\cdot3^6 + \binom{8}{6}\cdot2^6 - \binom{8}{7}$$

I exclude from the "universe" where 2 of my vertices are leaves.

2) My problem now is when I have at least 3 leaves, and using the same method $$\binom{8}{3}\cdot5^6 - \binom{8}{4}\cdot4^6 +\binom{8}{5}\cdot3^6 -\binom{8}{6}\cdot2^6 + \binom{8}{7}$$

Is everything true ? thank you !

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  • $\begingroup$ I am not sure I understand how you get your formula. $\endgroup$ – Arnaud Mortier Feb 10 '18 at 16:59
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By definition, a tree is connected, and it is a classical result that a tree with at least two vertices has at least two leaves.

Hence 1. is simply asking for the total number of trees on $8$ vertices, given by Cayley's formula:

$$8^6=262144$$

Now a tree has exactly two leaves if and only if it is linear, and such trees almost correspond to permutations on $8$ elements, the only caveat is that each tree corresponds to two permutations (since the trees are not supposed to be oriented). Hence there are $$\dfrac{8!}{2}$$ such trees, and so $$8^6-\dfrac{8!}{2}=241984$$ trees with more than $2$ leaves.

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