Example 2 Given $n(n \geqslant 6)$ points on a circle, connect each pair of points with a line segment, such that no three segments intersect at the same point inside the circle. Find the number of triangles determined by these segments whose vertices are inside the circle.

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Solution: Let the set of all triangles that meet the conditions be $A$, and $\triangle A_{1} A_{2} A_{3} \in A$. Extend the three sides of $\triangle A_{1} A_{2} A_{3}$ to intersect the circle at points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ (as shown in Figure 2). Let the set of all ways to choose 6 points from the $n$ given points be $B$, with $|B|=\mathrm{C}_{n}^{6}$. Then, the 6 points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ determine a selection $\alpha \in B$.
Establish a mapping $\varphi$ from $A$ to $B$:
For $\triangle A_{1} A_{2} A_{3} \in A$, let
$$
\varphi\left(\triangle A_{1} A_{2} A_{3}\right)=\alpha \text {. }
$$

It is easy to see that for different $\triangle A_{1} A_{2} A_{3} \in A$, the 6-point sets $\left\{B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}\right\}$ determined by the above method are also different, thus $\varphi\left(\triangle A_{1} A_{2} A_{3}\right)$ is also different, i.e., $\varphi$ is injective;

For any selection $\alpha \in B$, we can determine 6 points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ arranged in a counterclockwise order on the circle according to the selection $\alpha$. Connecting $B_{1} B_{4}, B_{2} B_{5}, B_{3} B_{6}$ yields $\triangle A_{1} A_{2} A_{3} \in A$, i.e., $\varphi\left(\triangle A_{1} A_{2} A_{3}\right)=\alpha$, hence $\varphi$ is surjective.
Therefore, $\varphi$ is bijective.
Thus, we have $|A|=|B|=\mathrm{C}_{n}^{6}$.