Example 11 There are 20 points distributed on a circle, and now we connect them using 10 chords that have no common endpoints and do not intersect each other. How many different ways can this be done?
(13th Moscow Olympiad problem)
Let $a_{n}$ denote the number of all different ways to connect $2n$ points on the circumference of a circle with $n$ non-intersecting chords. Considering the general case, it is obvious that $n=1$ is meaningless. We define $a_{0}=a_{1}=1$, and from the problem, we can derive the recurrence relation:
$$
a_{n}=a_{0} a_{n-1}+a_{1} a_{n-2}+a_{2} a_{n-3}+\cdots+a_{n-2} a_{1}+a_{n-1} a_{2}.
$$

Let the generating function of the sequence $\left\{a_{n}\right\}$ be $f(x)=\sum_{k=0}^{\infty} a_{k} x^{k}$. Since $a_{0}=a_{1}=1$, we have
$$
\begin{array}{l}
x \cdot f^{2}(x)=x \cdot\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+\cdots\right)^{2} \\
=a_{0}^{2} x+\left(a_{0} a_{1}+a_{1} a_{0}\right) x^{2}+\cdots+\left(a_{0} a_{n-1}+a_{1} a_{n-2}+\cdots+a_{n-1} a_{0}\right) x^{n}+\cdots \\
=a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+\cdots=f(x)-1,
\end{array}
$$

Thus, the generating function satisfies the equation $x \cdot f^{2}(x)-f(x)+1=0$, leading to $2 x \cdot f(x)=1 \pm \sqrt{1-4 x}$. Since $f(0)=a_{0}=1$, we have $f(x)=\frac{1}{2 x}(1-\sqrt{1-4 x})$.
Applying formula 5 to expand $(1-4 x)^{\frac{1}{2}}$ and substituting it into (*), we can obtain
$$
f(x)=\sum_{k=0}^{\infty} \frac{1}{k+1} C_{2 k}^{k} \cdot x^{k}, \text{ then } a_{n}=\frac{1}{n+1} C_{2 n}^{n}.
$$

Therefore, $a_{2}=2, a_{3}=5, a_{4}=14, a_{5}=42, \cdots, a_{10}=16796$.