Example 9 There are $n$ points on a circle $(n \geqslant 6)$, and a line segment is drawn between every two points. Suppose that any three segments do not intersect at a common point inside the circle, so any three segments intersect to form a triangle. Try to find the number of triangles determined by these segments.

The number of triangles formed by these segments.
We call the points on the circumference as outer points, and the intersection points of any two diagonals inside the circle as inner points. The triangles determined can be classified into 4 categories:
(1) The three vertices of the first type of triangle are all outer points, and the number of such triangles is denoted as $I_{1}$;
(2) The vertices of the second type of triangle include 2 outer points and 1 inner point, and the number of such triangles is denoted as $I_{2}$;
(3) The vertices of the third type of triangle include 1 outer point and 2 inner points, and the number of such triangles is denoted as $I_{3}$;
(4) The three vertices of the fourth type of triangle are all inner points, and the number of such triangles is denoted as $I_{4}$.
Obviously, the first type of triangle corresponds one-to-one with the set of 3 points on the circumference, so $I_{1}=C_{n}^{3}$.
Secondly, as shown in Figure 1-4(1), for any 4 points $A_{1}, A_{2}, A_{3}, A_{4}$ on the circumference, the line segments connecting them determine 4 second type triangles: $\triangle A_{1} O A_{2}, \triangle A_{2} O A_{3}, \triangle A_{3} O A_{4}, \triangle A_{4} O A_{1}$. Conversely, every 4 such second type triangles with a common inner vertex correspond to a 4-point set on the circumference, so $I_{2}=4 C_{n}^{4}$.

Similarly, as shown in Figure 1-4(2), for any 5 points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ on the circumference, the line segments connecting them determine 5 third type triangles: $\triangle A_{1} B_{1} B_{2}, \triangle A_{2} B_{2} B_{3}, \triangle A_{3} B_{3} B_{4}, \triangle A_{4} B_{4} B_{5}, \triangle A_{5} B_{5} B_{1}$, so we have $I_{3}=5 C_{n}^{5}$.

Finally, as shown in Figure 1-4(3), for any 6 points $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ on the circumference, they correspond to 1 fourth type triangle, so $I_{4}=C_{n}^{6}$.
In summary, the total number of triangles determined is $C_{n}^{3}+4 C_{n}^{4}+5 C_{n}^{5}+C_{n}^{6}$.
When we need to establish inequality relationships in combinatorial counting problems, we often construct injections or surjections to achieve this. Examples of this will be discussed in later chapters.