3-4. On a circle, there are 20 points. These 20 points are connected by 10 (non-intersecting - ed. note) chords. In how many ways can this be done?
Solution 4. Answer: 16796. Let \(a_{n}\) be the number of ways to connect \(2 n\) points on a circle with \(n\) non-intersecting chords. It is clear that \(a_{1}=1\) and \(a_{2}=2\). We will show that

\[
a_{n}=a_{n-1}+a_{n-2} a_{1}+a_{n-3} a_{2}+\cdots+a_{1} a_{n-2}+a_{n-1}.
\]

Fix one of the \(2 n\) points. The chord emanating from it divides the circle into two arcs, with an even number of given points on each arc: from 0 to \(2 n-2\). If there are \(2 k\) points on one arc, then there are \(2(n-k-1)\) points on the other arc; these points can be connected by non-intersecting chords (not intersecting the chord emanating from the fixed point) in \(a_{n-k-1} a_{k}\) ways.

Thus, \(a_{3}=a_{2}+a_{1}^{2}+a_{2}=5, a_{4}=14, a_{5}=42, a_{6}=132, a_{7}=429, a_{8}=1430, a_{9}=4862\) and \(a_{10}=16796\).

Remark. It can be proven that \(a_{n}=\frac{(2 n)!}{n!(n+1)!}=\frac{1}{2 n+1}\binom{2 n+1}{n}\).

\section*{9th - 10th Grades}