TITLE: How many vertices of degree 1 in a tree?
QUESTION [5 upvotes]: How many vertices of degree 1 are there in a tree with no vertices of degree more than 4?
The only thing that I have right now is that the number of edges in a tree is n-1 where n is the number of vertices. Any help would be appreciated. Thanks in advance.

REPLY [2 votes]: Let $k$ be the number of leafs in the tree. Let $n$ be the number of vertices.
It is known that $k \geq 2$ with equality for $T=P_n$.
As for the upper bound, the sum of degrees is at most $k+4(n-k)$. Thus 
$$2(n-1)\leq k+4(n-k)=4n-3k \,$$
Solving this equality we get 
$$k \leq \frac{2n+2}{3} \,.$$
Since $k$ is integer, we must have
$$k \leq \lfloor \frac{2n+2}{3} \rfloor \,.$$
The upper bound can be reached. Let $r$ be the remainder when $2n+2$ is divided by $3$. Then (as the above proof suggests) we need to construct a tree with $\lfloor \frac{2n+2}{3} \rfloor$ leafs, $r$ vertices of degree $2$ and $n-r- \lfloor \frac{2n+2}{3} \rfloor$ vertices of degree $4$. You can prove that such a tree exists by induction on $n$, the $P(n) \Rightarrow P(n+3)$ is easy (replace a leaf by a vertex of degree $4$ and connect it to 3 new leafs.)