Draw a planar graph with two vertices of degree 3 and four vertices of degree 5, if possible.

Question

Draw a planar graph with two vertices of degree 3 and four vertices of degree 5, if possible.

Attempt:

With handshaking lemma, I get this:

$2e = 26 \implies e=13$

Then with Euler's formula, I get:

$6-13+f=2 \implies f = 9$

However, since $e \leq 3v - 6$ for a simple, connected, planar graph I would get:

$13 \leq 3(6)-6$

$13 \leq 12$

I can't figure out how I could get 9 faces for my graph. I can only get 8 faces as shown here:

This is the best attempt I had on this problem with no success.

Answer 1

Here is a planar example. The vertices A and B are linked by two simple edges. The vertices A, B, E and F have degree 5. The vertices C and D have degree 3.

Answer 2

From formula you wrote $e\leq 3v-6$ you can see that such a (planar) graph does not exist.

Also you could note that this graph contains $K_{3,3}$ so again it can not be planar.

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Edit: Actualy this graph doesn't even exist since the sequence $5,5,5,5,3,3$ is not graphicaly. If it is, then following would be also

$$ 4,4,4,2,2\implies 3,3,1,1\implies 2,0,0$$ but last one clearly it is not graphicaly.