$n$ points are given on a circle. I'm looking for a formula for the number of pairs of disjoint chords connecting them. What I have so far:
I count the number of pairs of interior intersecting chords and subtract from the total number of pairs of chords on $n$ points. The latter can be calculated using: $\frac{\binom{n} {2}\binom{n-2}{2}}{2!}$
To count pairs of intersecting chords, I essentially go around the circle quadruple-counting the pairs (once for every point). Every chord determines two sides of the circle (half-planes) in which there are as few as 1 and as many as $n-3$ points for which points on one side can be connected with another point on the other side creating a pair of intersecting chords. In this manner, I get the following overcount:
$n\times(1\cdot(n-3)+2\cdot(n-4)+3\cdot(n-5)+\ldots+(n-3)\cdot1)$
Dividing this by 4 and subtracting from the total is my formula:
$\frac{\binom{n} {2}\binom{n-2}{2}}{2}-\frac{n\times(1\cdot(n-3)+2\cdot(n-4)+3\cdot(n-5)+\ldots+(n-3)\cdot1)}{4}$
Can anyone verify that my argument is valid?