A question on the restriction of the Euler’s formula We all know the famous Euler's Formula.
It says that if a polyhedron has F(Faces), V(vertices) and E(edges) then F + V – E = 2.
My question is “is there any restriction on these variables?”
By restriction, I mean something like aF + bV + cE > 0 (for some a, b and c) must be satisfied before the formula can be applied.
For example, if there is no such restriction, can I ask the following question:-
Given that E= 6 and V = 7, (i) find F; (ii) draw that figure and (iii) name that figure.
 A: (ii)/(iii): Putting the other parts of your question aside, this is something that is an issue. For instance, a decagonal prism and a dodecahedron both have 12 faces, 20 vertices, and 30 edges, so these counts alone don't always determine which polyhedron you have.
Finding F (or V or E) if there are any simple (no holes) polyhedra with the given counts for the other two numbers is what the formula is for. So if there were a polyhedron with 6 edges and 7 vertices, it must have only 1 face, but that wouldn't make for much of a polyhedron.
There are some basic inequalities to cut down what's possible, though. Since every edge has two halves, and every vertex is attached to at least three "half-edges", we have $2E\ge3V$ (your E=6 V=7 example doesn't obey this). Similarly, since every edge bounds exactly two faces, and every face has at least three edges, we have $2E\ge3F$. 
Edit: These inequalities, with Euler's formula, basically reduce to $(V+4)/2\le F\le2V-4$ and this table of polyhedron counts suggests that all numbers satisfying that are possible (they certainly are up to 32 vertices/faces).
A: Observe the following two constructions, where I assume $V$ is even.
Behold the glory of MS Paint.


In the first construction, $V/2$ vertices are on top, $V/2$ on the bottom, and there are just $\dfrac{3V}{2}$ edges, the minimal possible (as Mark S. points out).  In the second construction, I have added $\dfrac{V}{2}$ red and orange edges, and $V-6$ green edges, for a total of $3V-6$ edges, the maximal possible.  For any amount of edges in between, just choose any subset of the new edges in the second construction of the appropriate size.
The case where $V$ is odd I have not considered, but can probably be handled similarly. 
A: If the faces are triangles, you need to have $3F=2E$. Your example can't work, since a polyhedron cannot have just one face!
