Let G be a connected graph such that |E| = 17 and degv >= 3. Find max |V|?

Question

There would be 34 edges.

If we increase the vertices then we would decrease the Degree.

Hence 10 Vertices with degree of 3 and one with degree of 4.

So I think in total it would be 11 vertices am I right.

Answer 1

You are correct. To clarify some, you shouldn't say that if $E = 17$, then there are $34$ edges. You should note that if $E = 17$, then $2E = 34$. And so that's the right hand side for the handshaking argument.

So $3v + k = 34$. If $v = 10$, then $k = 4$. Since $\delta(v) = 3$ (the minimum degree), your other vertex must have degree $4$. Clearly $v \neq 11$, as that would imply the last vertex has degree $1$, a contradiction.