Number of quadrilaterals in a heptagon: is my reasoning correct? I found this question on a GRE prep site:

If you join all the vertices of a heptagon, how many quadrilaterals
  will you get?

There is a bunch of multiple choice answers but to me none of them seem correct and since it's not an official GRE site I am inclined to think that perhaps whoever made the question might have made a mistake.
First please let me explain how I understand the question:
Joining all the vertices to me means to draw the complete graph of the $7$-gon. A quadrilateral is a $4$-gon. So I have to find the number of $4$-gons inside a complete graph $K_7$.

I tried to count them as follows, please could someone tell me if this
  is correct or explain why it isn't?

I greatly appreciate your help since this is the very first question I've tried in order to prep for the GRE and it's making me extremely nervous that I can't seem to even understand the solution.
To count $4$-gons I need to first determine the total number of vertices -- that's the seven outer vertices plus all the new ones we get from intersecting lines inside. 
To do this I first determine the total number of lines and then determine with how many lines each lines intersects. So let's do that:
Starting with a random vertex we draw $7-3=4$ outgoing lines (minus three for the current vertex and the two adjacent ones). 
Moving to one of the two adjacent vertices we count $7-3=4$ outgoing lines. 
Now we move to the other adjacent vertex of the starting (=first) vertex. We have already drawn one outgoing line (coming from the second vertex we considered). So, the number of outgoing lines is $7-3-1=3$. 
Moving from this third vertex to the fourth (adjacent) vertex we note that we have already drawn two incoming lines into this one: one from the first and one from the second vertex we considered. So we count $7-3-2 = 2$ outgoing lines. 
We move on to the fifth vertex. By similar reasoning as before we count $7-3-3=1$ outgoing line. 
The sixth vertex will have $7-3-4=0$ outgoing lines. 
Altogether we drew $1+2+3+4+4=14$ lines. 
(Note: by now I realise that I could have used that this is the number of edges in a complete graph minus the 6 outside edges. So, somewhere I made a mistake since I should have $15$ lines drawn inside.).
Next we need to count how many points of intersection per line. So considering one particular line we note that it intersects with all lines that do not emanate from the same vertex. Per vertex there are $4$ emanating lines hence one particular line intersects with $15-4=11$ lines. 
Hence the number of intersection points is $11\cdot 15=165$.In addition to those we had $7$ vertices to begin with hence in total we have $165 + 7 = 172$ vertices. 
The number of $4$-gons to make from given $172$ vertices is 
$$ {172 \choose 4}$$
which seems a little large.
I also think that $165$ points of intersection in a heptagon seems too large.

I believe they intend your quadrilateral to have all of its vertices among the original $7$.  If you try to count the small quadrilaterals formed by pieces of the diagonals, it depends on where the original $7$ points are placed in the plane.  You can have multiple diagonals intersect at a point or not as you move the vertices around.  
If you don't mind quadrilaterals that have crossing sides, this is just the number of $4-$ cycles in $K_7$, answered here as ${7 \choose 4} \cdot 3=105$  If you do mind crossings, there is only one quadrilateral for each group of four points, so there are ${7 \choose 4}=35$
