Number of ways to separate $n$ points in the plane
The linked answer was to this question: Suppose you have $n$ points in the plane, no three of which are colinear. How do you show that the number of ways to separate them into two disjoint subsets by a line incident to none of them depends only on $n$ and not on the arrangement of the points?
Suppose we consider two lines equivalent if they both accomplish the same partition of the $n$ points. The proof given in the linked answer said we let such a line rotate counterclockwise until it bumps into two points and can rotate no further. This establishes a one-to-one correspondence between equivalence classes of lines and unordered pairs of points, whence we conclude that there are $\binom n 2$ equivalence classes; hence $\binom n 2$ ways to partition the points (not counting the trivial partition that puts all the points in one "part").
Now let us suppose that we had rotated clockwise rather than counterclockwise. Let us say of two equivalence classes $A$ and $B$ that $A\sim B$ iff the pair of points that stops the counterclockwise rotation of lines in $B$ is the same pair that stops the clockwise rotation of points in $A$.
Can anything of interest be said about this relation $A\sim B$?
For four points in the plane, there are six partitions of the kind described (not counting the trivial partition). Either of two alternatives must hold: (1) one of the four points is in the convex hull of the other three, or (2) none of them is in the convex hull of the other three. It is easy to see that the relation $A\sim B$ in the former case is not equivalent to that in the latter case.