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Could you find a 3-Regular Connected Planar Graph on $10$ vertices with $8$ faces? If so, explain carefully.

I don't know what does regular mean. I think that 3-connected graph on 10 vertices with 8 faces. From eulerian formula : $v + f - e = 2:$ $10+8-e=2 \Longrightarrow e=16$. And I use other proof "A simple planar graph on v 3 vertices has at most $\;3v- 6$ edges." $\Longrightarrow 30-6=24$ edges at most. I said that $24>16$, if so it can be.

Is it true or not?

Thank you for your answers in advance.

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    $\begingroup$ A $k$-regular graph is defined to be a graph where every vertex is of degree $k$. $\endgroup$
    – JMoravitz
    Commented Jan 20, 2015 at 16:01

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$3$-regular means that each vertex has degree $3$. With $10$ vertices, how many edges would that make?

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    $\begingroup$ 3*10/2 = 15 edges. Is that mean that it must be 16 from eular formula but we found 15. It is impossible. $\endgroup$
    – Zafer
    Commented Jan 20, 2015 at 16:05
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    $\begingroup$ Its all good. Already deleted that comment. So, now you know the number of faces, edges, and vertices. Check if it matches Euler's formula. $\endgroup$
    – JMoravitz
    Commented Jan 20, 2015 at 16:08

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