# Thread:

1. ## Simple graph proof question

Hi,

The exercise is as follows:

A simple graph, also called a strict graph, is an unweighted, undirected graph containing no graph loops or multiple edges. A well-known theorem states that the sum of the degrees of the vertices of a simple graph equals twice the number of edges of the graph.

Prove the following:

There is no simple graph with 12 vertices and 28 edges so that

(i) all vertices have degree 3 or 4
(ii) all vertices have degree 3 or 6
Thank you very much

2. Notice that $2(28)=56$.
So (i) is impossible.

For (ii) let s be the number of degree six vertices and t be the number of degree three.
Then $6s+3t=56~\&~s+t=12$.
What is wrong with that?
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