I wanted to know if my solution is correct. I am not sure if such questions are right for this site. Please let me know if this isn't.
Question: Is there a tree with 5 vertices of which two are of degree 3?
My Solution: Consider a tree with 5 vertices. Thus by theorem (which states that a tree on n vertices has n-1 edges), it should have 5-1=4 edges.
By handshaking lemma, total degree of the tree (sum of degrees of all vertices) is 2$\times$ (No. of edges) = 2$\times$4 =8
This tree has 2 vertices each of degree 3 $\implies$ The sum of degrees of remaining 3 vertices is 8-3-3=2
Now, it is impossible to have 3 vertices in a tree with sum of degrees = 2 without having at least one of the vertices have degree zero. (I think this line in my proof isn't rigorous enough)
But if at least one of the vertices have degree zero, the graph cannot be a tree as it becomes disconnected (A tree by definition is a connected acyclic undirected graph)
Does this proof make sense?