Example 2.2.5 There are $n$ points on a circle $(n \geqslant 6)$, and each pair of points is connected by a line segment. Assuming that no three segments intersect at the same point inside the circle, three segments that intersect each other pairwise form a triangle. Try to find the number of these triangles.
Analysis: Let the vertices of the triangle be $i$ inside the circle and $3-i$ on the circumference. Let the set of all such triangles be $I_{i}(i=0,1,2,3)$, and find $\left|I_{i}\right|$, then use the principle of addition.
Solution: $\left|I_{0}\right|=\mathrm{C}_{n}^{3}$.
To calculate $\left|I_{1}\right|$, for $\triangle \in I_{1}$, there is an inner point $O$ which is a vertex of $\triangle$, and the inner point $O$ is the intersection of two line segments, corresponding to four points $A_{1}, A_{2}, A_{3}, A_{4}$ on the circumference, where $A_{1} A_{3} \times A_{2} A_{4}=0$. That is, there is a one-to-one correspondence between the set of all inner points and the set of four points on the circumference, and for each inner point $O$, there are four $\triangle$ in $I_{1}$ corresponding to it, hence $\left|I_{1}\right|=4 \mathrm{C}_{n}^{4}$.

Next, calculate $\left|I_{2}\right|$, by choosing any 5 points from the $n$ points on the circumference, there correspond 5 $\triangle$ in $I_{2}$; conversely, for each $\triangle \in I_{2}$, extending the sides of $\triangle$ to intersect the circumference at 5 points, making this $\triangle$ one of the 5 $\triangle$ corresponding to the 5 points, hence $\left|I_{2}\right|=5 \mathrm{C}_{n}^{5}$.

For $\triangle \in I_{3}$, the three inner points determine three line segments intersecting the circumference at 6 points; conversely, the same applies, hence $\left|I_{3}\right|=\mathrm{C}_{n}^{6}$.
Therefore, the total number of triangles $=\mathrm{C}_{n}^{3}+4 \mathrm{C}_{n}^{4}+5 \mathrm{C}_{n}^{5}+\mathrm{C}_{n}^{6}$.