TITLE: Every two Strings that don't cross each other on a circle
QUESTION [1 upvotes]: here a picture of a circle with n points on it.

I need to count how many two strings "groups" on that circle, don't cross each other. to be more precise, "every two strings, need NOT to cross each other.." im bad at english, sorry.
Btw, on my opinion:$ C(n,4)$ is the first part of the answer. (like you see on the picture, we took a set of 4 points of n, but it is also the number of squares that you can build - to be more precise, convex squares, which means the answer for this question is slighly different) or some sort of $C(n,2) + 'something'$
But I believe I need to add something here. Do you guys have any ideas?

REPLY [1 votes]: We assume the points are in general position. It is not clear whether by two-string pairs you mean ordered pairs or unordered. We use unordered.
We interpret "don't cross" as meaning that the pair don't cross in the interior of the circle. So if $A,B,C$ are distinct points, then we say that $AB$ and $AC$ don't cross.     
Take $4$ distinct points on the circle. These determine $3$ unordered pairs of strings, of which $2$ pairs do not meet in the interior of the circle.
Thus the probability two unordered pairs don't cross is $\frac{2}{3}$.  There are $\frac{1}{2}\cdot\binom{n}{2}\cdot \binom{n-2}{2}$ unordered pairs of strings. For the number that don't cross, multiply by $\frac{2}{3}$.
If we consider $AB$ and $AC$ as crossing, multiply by $\frac{1}{3}$ instead of $\frac{2}{3}$.
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