A polyhedron has $F$ faces. Find all possible amounts of its vertices and edges.
Of course, for small values of $F$, it's possible to just list the possibilities. But how to handle the general form of that task? Just plugging the faces into Euler's formula won't be enough.
In such a question, as the cited one, "polyhedron" usually is meant to be defined as convex, i.e. the set of points contained within the intersection of halfspaces. There Euler's formula does apply in its easiest form, i.e. according to $V-E+F=2$. $-$ Thence in the followings we assume this setting to be given.
First we consider the medium number $S$ of edges per vertex. That one then obeys the following equation: $VS=2E$. And furthermore, by geometric restriction, we can assume $S\ge 3$. In other words we get $3\le S=2E/V$.
Solving this relation for $V$ and inserting that into Euler's equation finally provides $E\le3F-6$. Solving it instead for $E$ and inserting that into Euler's equation instead would provide $V\le2F-4$.
Next we consider the medium number $N$ of edges per face. That one then obeys the following equation: $NF=2E$. And furthermore, by geometric restriction, we can assume $N\ge 3$. In other words we get $3\le N=2E/F$.
Solving this relation for $E$ directly provides $3F/2\le E$. Solving instead for $F$ and inserting that into Euler's equation finally provides $F/2+2\le V$.
Thus, combining these results, we get: $$\begin{array}{rcccl} F/2+2 & \le & V & \le & 2F-4\\ 3F/2 & \le & E & \le & 3F-6 \end{array}$$
As already was mentioned within a comment, these ranges however do not tell whether any according value pair of $V$ and $E$ truely would be taken, i.e. that there would exist an according polyhedron with such indivdual values. It just tells the opposite: outside of these ranges there cannot be found any (convex) polyhedron with such values.
--- rk