Is there any close form solution to this question? For $$N=4$$, it is just $$p^4$$, but $$N=5$$ we have to consider all the possibilities from 1~5 Quadrilaterals.
• That helps, thank you. But in light of what you have written, it seems to me the answer must depend on the specific configuration of the points. Even with $N=4$ the answer can be $3p^4$ or $p^4$ depending on whether the convex hull of the points is a triangle or a quadrilateral, respectively. What, then, are you assuming about the point configuration that would enable a unique answer to be provided? – whuber Mar 13 '19 at 19:20
• Assuming all convex hulls of all subsets of four points are quadrilaterals is a severe restriction: it implies the configuration is essentially that of a sequence of $N$ points placed on a circle. – whuber Mar 13 '19 at 19:52