Example 4 In a convex $n$-sided polygon, no three diagonals intersect at a single point inside the polygon. Questions: (1) How many regions do these diagonals divide the convex $n$-sided polygon into? (2) How many segments do these diagonals get divided into by their intersection points?
(1) Let the convex $n$-sided polygon be divided into regions by its diagonals, with the region having the most sides being an $m$-sided polygon. Among these, there are $n_{3}$ triangles, $n_{4}$ quadrilaterals, ..., and $n_{m}$ $m$-sided polygons. Thus, the number of regions into which the convex $n$-sided polygon is divided is
$$
S=n_{3}+n_{4}+\cdots+n_{m} \text {. }
$$

On one hand, the total number of vertices of all regions is $3 n_{3}+4 n_{4}+\cdots+m \cdot n_{m}$.
On the other hand, there are $C_{n}^{4}$ intersection points of the diagonals inside the polygon, and each intersection point is a common vertex of 4 regions (since no 3 diagonals intersect at a single point inside the polygon). Additionally, the convex $n$-sided polygon has $n$ vertices, each of which is a common vertex of $n-2$ regions, so
$$
3 n_{3}+4 n_{4}+\cdots+m \cdot n_{m}=4 C_{n}^{4}+n(n-2) .
$$

Furthermore, the sum of the interior angles of all regions is $n_{3} \cdot 180^{\circ}+n_{4} \cdot 360^{\circ}+\cdots+n_{m}(m-2) \cdot 180^{\circ}$.
On the other hand, the sum of the interior angles at the $n$ vertices of the original convex polygon is $(n-2) \cdot 180^{\circ}$, and the sum of the interior angles at the $C_{n}^{4}$ intersection points of the diagonals is $C_{n}^{4} \cdot 360^{\circ}$, so
$$
\begin{array}{l}
n_{3} \cdot 180^{\circ}+n_{4} \cdot 360^{\circ}+\cdots+n_{m} \cdot(m-2) \cdot 180^{\circ}=360^{\circ} \cdot C_{n}^{4}+180^{\circ}(n-2), \text { i.e., } \\
n_{3}+2 n_{4}+\cdots+(m-2) n_{m}=2 \cdot C_{n}^{4}+(n-2) .
\end{array}
$$
Subtracting (2) from (1), we get
$$
\begin{aligned}
S=n_{3}+n_{4}+\cdots+n_{m} & =C_{n}^{4}+\frac{1}{2}(n-1)(n-2)=C_{n}^{4}+C_{n-1}^{2} \\
& =\frac{1}{24}(n-1)(n-2)\left(n^{2}-3 n+12\right) .
\end{aligned}
$$
(2) From each intersection point inside the polygon, there are 4 line segments on the diagonals, and there are $C_{n}^{4}$ intersection points inside the polygon, making a total of $4 \cdot C_{n}^{4}$ line segments. Additionally, from each vertex of the convex $n$-sided polygon, there are $n-3$ line segments on the diagonals, so from the $n$ vertices, there are $n(n-3)$ line segments: but each line segment is counted twice in the above count, so the number of line segments into which the diagonals are divided by the intersection points is
$$
\frac{1}{2}\left[4 C_{n}^{4}+n(n-3)\right]=2 C_{n}^{4}+\frac{1}{2} n(n-3)=\frac{1}{12} n(n-3)\left(n^{2}-3 n+8\right) .
$$