Example 43 (3rd Moscow Mathematical Olympiad, 1990 National Training Team Problem, 1991 Canadian Mathematical Olympiad) In a convex $n$-sided polygon, no three diagonals intersect at a point inside the polygon. How many regions do these diagonals divide the convex $n$-sided polygon into?
Solution 1 Let the convex $n$-sided polygon be divided into regions, among which the convex polygon with the most sides is an $m$-sided polygon. Let the number of $k$-sided regions be $n_{k}$ $(k=3,4, \cdots, m)$. Thus,

On one hand, the total sum of the interior angles of each polygon 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 each intersection point inside the shape is $360^{\circ}$, and the total number of intersection points of the diagonals inside the shape is $C_{n}^{4}$ (the conclusion of Example 8 in the counting problem), while the sum of the interior angles at each vertex of the original convex $n$-sided polygon is $(n-2) \cdot 180^{\circ}$. Therefore, the total sum of the interior angles of each region is
$$
C_{n}^{4} \cdot 360^{\circ}+(n-2) \cdot 180^{\circ}.
$$

From (1) and (2), we get
$$
\begin{array}{l}
n_{3} \cdot 180^{\circ}+n_{4} \cdot 360^{\circ}+\cdots+n_{m}(m-2) \cdot 180^{\circ}=C_{n}^{4} \cdot 360^{\circ}+(n-2) \cdot 180^{\circ}, \text { i.e., } \\
n_{3}+2 n_{4}+\cdots+(m-2) \cdot n_{m}=2 C_{n}^{4}+(n-2).
\end{array}
$$

The total number of vertices of each region is
$$
3 n_{3}+4 n_{4}+\cdots+m n_{m}.
$$

On the other hand, there are $C_{n}^{4}$ intersection points inside the shape, and each intersection point is a common intersection point of 4 regions (since no three diagonals intersect at one point inside the shape). Also, each vertex of the convex $n$-sided polygon is a common intersection point of $n-2$ regions, so the total number of vertices of each region is
$$
4 C_{n}^{4}+n(n-2).
$$

From (4) and (5), we get
$$
3 n_{3}+4 n_{4}+\cdots+m n_{m}=4 C_{n}^{4}+n(n-2).
$$
Subtracting (3) divided by 2 from (6) yields $n_{3}+n_{4}+\cdots+n_{m}=C_{n}^{4}+C_{n-1}^{2}$,

which means the number of polygonal regions into which the convex $n$-sided polygon is divided is $C_{n}^{4}+C_{n-1}^{2}$.
Solution 2 Because removing each diagonal reduces the number of regions by $a_{i}+1$. Here, $a_{i}$ is the number of intersection points of the $i$-th diagonal removed with the other diagonals that have not yet been removed. Gradually removing $C_{n}^{2}-n$ diagonals, we are left with only one region, so the number of regions sought is $S=1+\sum_{i=1}^{C_{n}^{2}-n}\left(a_{i}+1\right)$, and $\sum_{i=1}^{2_{n}^{2}-n} a_{i}$ is exactly the total number of intersection points of the diagonals inside the shape $C_{n}^{4}$, so the number of regions sought is
$$
S=1+C_{n}^{1}+C_{n}^{2}-n=C_{n}^{4}+C_{n-1}^{2}.
$$

Solution 3 Let the number of regions into which the convex $n$-sided polygon is divided by diagonals be $a_{n}$, then $a_{3}=1, a_{4}=4$.
Starting from the convex $(n-1)$-sided polygon $A_{1} A_{2} \cdots A_{n-1}$ and adding the vertex $A_{n}$, we get the convex $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, thus adding a $\triangle A_{1} A_{n-1} A_{n}$. Then connecting $A_{n} A_{2}, A_{n} A_{3}, \cdots, A_{n} A_{n-2}$, these $n-3$ diagonals add $C_{n}^{4}-C_{n-1}^{4}$ intersection points, which divide the $n-3$ diagonals into $C_{n}^{4}-C_{n-1}^{4}+(n-3)$ segments, each segment dividing the original region into two, so it adds $C_{n}^{4}-C_{n-1}^{4}+(n-3)$ regions, hence
$$
a_{n}-a_{n-1}=1+C_{n}^{4}-C_{n-1}^{4}+(n-3)=C_{n}^{4}-C_{n-1}^{4}+(n-2).
$$

Defining $a_{2}=0, C_{2}^{4}=C_{3}^{4}=0$, then (*) holds for all $n \geqslant 3$. Therefore, the number of regions into which the convex $n$-sided polygon is divided is
$$
\begin{aligned}
a_{n} & =a_{2}+\sum_{k=3}^{n}\left(a_{k}-a_{k-1}\right)=0+\sum_{k=3}^{n}\left(C_{k}^{1}-C_{k-1}^{4}\right)+\sum_{k=3}^{n}(k-2) \\
& =C_{n}^{4}+\frac{1}{2}[1+(n-2)](n-2)=C_{n}^{4}+C_{n-1}^{2}.
\end{aligned}
$$

Solution 4 Let the number of regions into which the convex $n$-sided polygon is divided be $S$, then the number of regions into which the entire plane is divided is $F=S+1$, the number of vertices (including the vertices of the convex $n$-sided polygon and the intersection points of the diagonals) is $V=C_{n}^{4}+n$, and the number of edges is $E=2 C_{n}^{4}+$ $\frac{1}{2} n \cdot(n-1)=2 C_{n}^{1}+C_{n-1}^{2}$ (because from each vertex of the convex $n$-sided polygon, there are $n-1$ edges, and from each intersection point of the diagonals inside the shape, there are 4 edges, so the total number of edges from all vertices is $n(n-1)+4 C_{n}^{4}=2 C_{n}^{2}+4 C_{n}^{4}$, but in the above count, each edge is counted twice, so the number of edges sought is $\left.E=\frac{1}{2}\left(4 C_{n}^{4}+2 C_{n}^{2}\right)=2 C_{n}^{4}+C_{n}^{2}\right)$, substituting into Euler's formula $V+F-E=3$, we get
$$
S=E-V+1=2 C_{n}^{4}+C_{n}^{2}-\left(C_{n}^{4}+n\right)+1=C_{n}^{4}+C_{n-1}^{2}.
$$