[ Systems of points and segments (other) ] [ Classical combinatorics (other) ]

On a circle, 20 points are marked. How many triples of chords with endpoints at these points exist such that each chord intersects the other two (possibly at the endpoints)?
The ends of the chords in a triplet can be 3, 4, 5, or 6 points. Let's consider these cases. 1) The ends of the chords are 3 points (see fig.). They can be chosen in $C_{20}^{3}$ ways. Each triplet of points can be connected by chords in a unique way.

![](https://cdn.mathpix.com/cropped/2024_05_06_9bfea73962495faf8ed9g-24.jpg?height=343&width=458&top_left_y=66&top_left_x=800)

2) The ends of the chords are 4 points. There are two possible configurations of the chords (see fig.). Four points can be chosen in $C_{20}^{4}$ ways. For each set of four points, there are 8 ways to connect them with chords.

![](https://cdn.mathpix.com/cropped/2024_05_06_9bfea73962495faf8ed9g-24.jpg?height=414&width=886&top_left_y=576&top_left_x=586)

3) The ends of the chords are 5 points. In this case, exactly two chords share a common vertex, and the third chord connects the remaining two points (see fig.). Five points can be chosen in $C_{20}^{5}$ ways. For each set of five points, there are five ways to draw the chords (one for each point where two chords meet).

![](https://cdn.mathpix.com/cropped/2024_05_06_9bfea73962495faf8ed9g-24.jpg?height=417&width=377&top_left_y=1161&top_left_x=838)

4) The ends of the chords are 6 points (see fig.). Six points can be chosen in $C_{20}^{6}$ ways. For each set of six points, there is a unique way to draw the chords, as the chords must intersect pairwise at internal points.

![](https://cdn.mathpix.com/cropped/2024_05_06_9bfea73962495faf8ed9g-24.jpg?height=388&width=440&top_left_y=1764&top_left_x=818)

Thus, the total number of ways to draw the chords is $C_{20}^{3} + C_{20}^{4} \cdot 8 + C_{20}^{5} \cdot 5 + C_{20}^{6} = 156180$.

## Answer

156180.