There are N points on a circle. If we draw all the chords connecting these points and no three of them intersect at the same point, in how many parts will the interior of the circle get broken?
For example, when N is equal to 1, 2, 3, 4, and 5, we get 1, 2, 4, 8, and 16 parts respectively.
The answer, somewhat surprisingly, is not 2ᴺ⁻¹, but 1 + N(N-1)/2 + N(N-1)(N-2)(N-3)/24.
In order to see that, we start with a single sector, the interior of the circle, and keep successively drawing chords. Every time we draw a new chord, we increase the number of parts by 1 and then add 1 extra part for each intersection with previously drawn chords.
Therefore, the total number of parts at the end will be:
1 + the number of the chords + the number of the intersections of the chords
Each chord is determined by its 2 endpoints and therefore the number of chords is N(N-1)/2.
Each intersection is determined by the 4 endpoints of the two intersecting chords and therefore the number of intersections is N(N-1)(N-2)(N-3)/4!.