How to determine the number of the non self-intersecting quadrilaterals that can be formed from 4 points? Basically, I'm trying to find the requirements needed to determine whether one or three non self-intersecting quadrilaterals can be formed from 4 given points. 
 A: Consider the complete graph $\Gamma$ on your four vertices, which looks like the shadow of a tetrahedron. A quadrilateral is obtained by deleting two edges that do not share a vertex (i.e., making a circuit on the shadow of the $1$-skeleton of the tetrahedron. If no three of your vertices are collinear, there are two cases:


*

*The points are vertices of a convex quadrilateral (i.e., two edges of $\Gamma$ cross), and the only "non-self-intersecting" quadrilateral is the boundary of the convex hull (the crossing edges are the only pair that can be removed).

*One vertex is inside the triangle formed by the remaining three (i.e., no two edges of $\Gamma$ cross), and there are three (non-convex) quadrilaterals, obtained by removing an edge inside the triangle and the "complementary" edge of the triangle.
If three vertices are collinear, I'd like to see a precise definition of "non-crossing quadrilateral" before attempting to enumerate cases. (Perhaps I've misunderstood your question, since there don't seem to be any cases in which exactly two quadrilaterals arise.)
