Example 12 There are $n$ points on a circle $(n \geqslant 6)$, and each pair of points is connected by a line segment. Assuming that any 3 line segments do not intersect at a common point inside the circle, thus any three intersecting line segments form a triangle. Try to find the number of triangles determined by the connected line segments.

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The so-called fixed triangles can be divided into 4 categories $S_{0}, S_{1}, S_{2}, S_{3}$, where the triangles in $S_{i}$ have exactly $i$ vertices on the circumference of the circle, and the remaining $3-i$ vertices are inside the circle $(i=0,1,2,3)$.
$S_{3}$ triangles correspond one-to-one with sets of 3 points on the circumference, so $\left|S_{3}\right|=C_{n}^{3}$.
As shown in Figure 11-2(1), a set of 4 points on the circumference $A_{1}, A_{2}, A_{3}, A_{4}$ determines 4 triangles in $S_{2}$: $\triangle O A_{1} A_{2}$, $\triangle O A_{2} A_{3}$, $\triangle O A_{3} A_{4}$, $\triangle O A_{4} A_{1}$. Conversely, such 4 triangles in $S_{2}$ with a common vertex $O$ correspond to a set of 4 points on the circumference, so $\left|S_{2}\right|=4 C_{n}^{4}$.

Similarly, from Figure 11-2(2) and Figure 11-2(3), we get $\left|S_{1}\right|=5 C_{n}^{5},\left|S_{0}\right|=C_{n}^{6}$, hence the total number of triangles determined is $\left|S_{3}\right|+\left|S_{2}\right|+\left|S_{1}\right|+\left|S_{0}\right|=C_{n}^{3}+4 C_{n}^{4}+5 C_{n}^{5}+C_{n}^{6}$.
Note: This problem is a typical example of solving a problem using a one-to-many correspondence method.
(3) The method of counting twice
The principle of counting twice (also known as Fubini's principle) states that for the same quantity, if it is calculated in two different ways, the results should be equal. Of course, if one calculation is precise and the other is an estimate (only calculating an upper or lower bound), then an inequality will be obtained.
This method of solving problems by establishing equations or inequalities is called the method of counting twice.