i have a simple planar graph $G$ with $6$ vertices,$4$ vertices of degree $4$. Why $G$ doesn't have a vertex with degree of $5$?
I know it's suppose to be a simple question but IM clueless and cant find a way to deal with it ,I tried to show that $G$ contains $K_5$ or $K_{3,3}$ graph but without any success so far ,thanks to all the helpers
Suppose the graph has a vertex of degree $5$.
Then it is connected so it is true that
$$e \le 3v-6 \implies e \leq 12$$
but $$\sum_{v \in V} deg(v) =5+4\cdot 4+x = 21+x = 2e$$ where $x$ is the degree of the vertex with not specified degree, then $x = 1$ or $x=3$ .
If $x=1$ then this means the four vertices of degree $4$ have to form $K_4$ and it's easy to see that $K_4$ with all the four vertices connected to another vertex is $K_{5}$.
Instead if $x=3$ you can easily draw the graph and see that all choices are bound and you will obtain again a supergraph of (an expansion of) $K_{5}$