Show that there is no simple graph with four vertices such that three vertices have degree 3 and one vertex has degree 1 [closed]
I know how to represent this with a graph, but I would like to explain it with theory.
P.S. The question is from "Introduction to Graph theory" by Clark & Holton.
Solution 1:
to continue with @Globos answer, after removing a vertex of degree 1, the graph has 3 vertices left, with degree sequence 3,3,2. by the handshake lemma, the number of edges = (3+3+2)/2 = 4. but a simple graph with 3 vertices can only have 3 edges at most, contradiction.