Forming Combinations of Quadrilaterals From Heptagon What is the formula for finding Number of  Quadrilaterals from heptagon or any other regular polygon above pentagon for that matter? 
 A: As pointed out in the comments, a much simpler way is to just count the ways to select any four of the seven (or whatever) vertices.   $\binom 7 4$ for a heptagon, or $\binom n 4$ for any polygon of $n$ sides/vertices.
The following is overly complex, and I really don't know what I was thinking at the time.

For a heptagon there are 7 vertices.  That gives seven starting points, label them $A,B,C,D,E,F,G$, moving clockwise.  
We need to account for rotational symmetry.  If we choose any four vertice $WXYZ$, that is rotationally equivalent to three others: $XYZW$, $YZWX$, and $ZWXY$.  The only difference being from were we start.
If we select any of the seven as a starting point, say $A$ for convenience, there are then four points, moving clockwise, we can use for the diagonally-opposite vertix of the quadrilateral. That's $C,D,E,F$.  Now as $C$ is two vertices clockwise from $A$, and $F$ is five; that's equivalent to choosing a number, $y \in [2..5]$.
If we choose $C$ then the remaining two points on our quadrilateral have to be $B$ and a choice of $D,E,F,G$.  If we choose $D$ then the other two are a choice of $B,C$ and a choice of $E,F,G$.  Et cetera.  Again this is simply choosing numbers of vertices clockwise from $A$, which are $x\in [1..y-1]$ and $z \in [y+1..6]$
So we count by the summation: $$C(7) = \frac{7}{4} \sum_{y = 2}^4 \sum_{x = 1}^{y-1} \sum_{z=y+1}^6 1 = \frac{7}{4} \sum_{y = 2}^4 (y-1)(6-y) = 35$$
Thus there are 35 distinct quadrilaterals that are can be formed by selecting 4 vertices of a heptagon.

So in general you wish to find: $C(n) = \frac n 4 \sum\limits_{y=2}^{n-2} (y-1)(n-1-y)$ 
Reindexing, that is: $C(n) = \frac 4 2 \sum\limits_{k=1}^{n-3} ((n-2)k - k^2)$
Applying the formula for $\sum\limits_{k=1}^N k = \frac{N(N+1)}2$ and $\sum\limits_{k=1}^N k^2 = \frac{N(N+1)(2N+1)}{6}$, we have:
$$C(n) = \frac{n}{4}\left((n-2)\frac{(n-3)(n-2))}{2}-\frac{(n-3)(n-2)(2n-5)}{6}\right)$$
Which simplifies to: $$C(n) = \frac{n(n-1)(n-2)(n-3)}{24}$$

Pentagon: $C(5) = 5$, Hexagon: $C(6)=15$, Heptagon: $C(7)=35$, Octagon: $C(8)=70$, and so on.
