$1 \cdot 91$ Let there be $n$ points on a circle $(n \geqslant 6)$, where each pair of points is connected by a chord and no three chords have a common point inside the circle. How many different triangles can be formed by the intersections of these chords?
[Solution] (1) First, consider the set $S_{1}$ of triangles with all 3 vertices on the circle. Let $T_{1}$ be the set of all 3-element subsets of the $n$ known points on the circle, then $\left|T_{1}\right|=C_{n}^{3}$. Clearly, the correspondence from a triangle to its 3 vertices is a bijection from $S_{1}$ to $T_{1}$, hence $\left|S_{1}\right|=C_{n}^{3}$.
(2) Next, consider the set $S_{2}$ of triangles with 2 vertices on the circle and 1 vertex inside the circle. Let $T_{2}$ be the set of all 4-element subsets of the $n$ known points, then $\left|T_{2}\right|=C_{n}^{4}$.

Let $\triangle M P_{i} P_{j}$ be any such triangle, with vertices $P_{i}$, $P_{j}$ being known points on the circle, and vertex $M$ being the intersection of chords $P_{i} P_{i}^{\prime}$ and $P_{j} P_{j}^{\prime}$ inside the circle. Let $\triangle M P_{i} P_{j}$ correspond to the set of four points $\left\{P_{i}, P_{j}, P_{i}^{\prime}, P_{j}^{\prime}\right\}$, then $\triangle M P_{j} P_{i}^{\prime}$, $\triangle M P_{i}^{\prime} P_{j}^{\prime}$, and $\triangle M P_{j}^{\prime} P_{i}$ also correspond to this same set of four points. It is clear that this mapping is a multiple mapping from $S_{2}$ to $T_{2}$ with a multiplicity of 4. Thus, $\left|S_{2}\right|=4 C_{n}^{4}$.
(3) Then, consider the set $S_{3}$ of triangles with 1 vertex on the circle and 2 vertices inside the circle. Let $T_{3}$ be the set of all 5-element subsets of the $n$ known points, then $\left|T_{3}\right|=C_{n}^{5}$.

Let $\triangle P_{1} Q_{1} Q_{5}$ be such a triangle, with vertex $P_{1}$ on the circle and vertices $Q_{1}$ and $Q_{5}$ inside the circle (as shown in the figure). Let $\triangle P_{1} Q_{1} Q_{5}$ correspond to the set of five points $\left\{P_{1}, P_{2}, P_{3}, P_{4}, P_{5}\right\} \in T_{3}$, which defines a multiple mapping from $S_{3}$ to $T_{3}$ with a multiplicity of 5. Hence, $\left|S_{3}\right|=5\left|T_{3}\right|=5 C_{n}^{5}$.
(4) Finally, consider the set $S_{4}$ of triangles with all 3 vertices inside the circle. Let $T_{4}$ be the set of all 6-element subsets of the $n$ known points, then $\left|T_{4}\right|=C_{n}^{6}$.

Let $\triangle M_{1} M_{2} M_{3}$ be such a triangle (as shown in the figure). Let $\triangle M_{1} M_{2} M_{3}$ correspond to the set of six points $\left\{P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}\right\} \in T_{4}$. It is clear that this is a bijection from $S_{4}$ to $T_{4}$. Thus, $\left|S_{4}\right|=\left|T_{4}\right|=C_{n}^{6}$.
By the principle of addition, the total number of triangles is $S=C_{n}^{3}+4 C_{n}^{4}+5 C_{n}^{5}+C_{n}^{6}$.