Which could be a possible time complexity?
$n=|V|$, and $\sum_{v\in V}deg(v) = 2\cdot |E| = 2m$ (each edge is counted exactly twice).
Take a look here for a proof why ${n \choose k} = \Theta(n^k)$ for a fixed $k$. Substitute everything, and get that also $O(n^k*m)$ is a valid estimate (again, only if $k$ is a fixed number!).
c) Is clearly not a tight bound, since it doesn't even depend on $k$ (although, this statement is by no means a proof).