Question: Let G be a simple graph with 6 vertices and 10 edges such that every vertex of G has an odd degree. If the number of vertices of degree 3 is one more that the number of vertices of degree 5, how many vertices of each degree does G have?
I don't seem to have anything in my notes, but my thought process right now is that if we have a vertex of degree 3 I must have 2 vertices of degree 5, which sum up to 13. Since I'm given the edges, and according to the handshake lemma it needs to equal to 20. Where do I go from here.