Suppose there are k points, no 3 of which are collinear. What is the upper bound on the number of quadrilaterals we can form? We understand that the number of triangles possible is given by kC3, since any selection of 3 points uniquely determine a triangle.
This is not true for quadrilaterals though, since for selections where 1 of the points lie within the other 3, we can obtain 3 quadrilaterals. The value kC4 merely gives us a lower bound, which occurs whenever all k points lie on a circle (or whenever none of the k points lie within any possible triangle, but this condition is guaranteed by a placing them on a circle).
Now, my question is: what is the upper bound on the number of quadrilaterals we may form? Under what configuration of the k points will this value be the actual answer? Even more generally, is there an explicit formula in k and n giving us the upper bound for an n-gon and is there a rule which will help us determine how these k points should be placed?
Thank you in advance.
 A: Your question is equivalent to determining the rectilinear crossing number of the complete graph on $k$ vertices. This is an open problem! I don't know much about it, so I'll refer you to a survey article from 2011: http://www.csun.edu/~ba70714/survey.pdf
Edit: Here's the explicit connection. For every four vertices, consider the three pairs of rectilinear edges between them. Either exactly one pair crosses, in which case there is only one simple quad that passes through the vertices, or no pairs cross, in which case there are three simple quads that pass through the vertices. So if there are $c$ crossings, then the number of simple quads is
$$3\binom k4-2c.$$
For example, the rectilinear crossing number of $K_6$ is $3$, so the maximum number of simple quads formed by $6$ points is
$$3\binom 64-2\times3 = 45-6=39.$$
Here's a drawing of $K_6$ with only $3$ crossings, which I stole from this PDF:

I guess if you're patient enough, you can actually count $39$ simple quads in there. I haven't tried this myself. :)
