1.94 Let a convex $n(n \geqslant 4)$-gon be such that no three diagonals are concurrent. How many non-overlapping smaller polygons does this $n$-gon get divided into by all its diagonals?
[Solution 1] Let the convex $n$-sided polygon be divided into regions, with the polygon having the most sides being an $m$-sided polygon, and let the number of $k$-sided polygons be $n_{k}, k=3,4, \cdots, m$. We will calculate the total number of vertices and the sum of the interior angles from two perspectives.

On one hand, the total number of vertices of the divided polygons is
$$
3 n_{3}+4 n_{4}+\cdots+m n_{m} \text {, }
$$

and the total sum of the interior angles of the divided polygons is
$$
n_{3} \cdot 180^{\circ}+2 n_{4} \cdot 180^{\circ}+\cdots+(m-2) n_{m} \cdot 180^{\circ} .
$$

On the other hand, each vertex of the original $n$-sided polygon is a common vertex of $n-2$ small polygons, and any intersection of two diagonals is a vertex of 4 small polygons, with such intersections totaling $C_{n}^{4}$. Therefore, the total number of vertices of the divided polygons should also be $n(n-2)+4 C_{n}^{4}$. From this and (1), we get
$$
3 n_{3}+4 n_{4}+\cdots+m n_{m}=n(n-2)+4 C_{n}^{4} .
$$

Moreover, the sum of the interior angles of the original $n$-sided polygon is $(n-2) 180^{\circ}$, and each intersection of diagonals contributes $360^{\circ}$ to the sum of the interior angles. Therefore, the total sum of the interior angles of the divided polygons should also be $(n-2) 180^{\circ}+$ $C_{n}^{4} 360^{\circ}$. From this and (2), we get
$$
n_{3}+2 n_{4}+\cdots+(m-2) n_{m}=n-2+2 C_{n}^{4} \text {. }
$$
Subtracting (4) from (3) gives
$$
\begin{array}{l}
2\left(n_{3}+n_{4}+\cdots+n_{m}\right)=(n-1)(n-2)+2 C_{n}^{4}, \\
n_{3}+n_{4}+\cdots+n_{m}=C_{n-1}^{2}+C_{n}^{4},
\end{array}
$$

which means the total number of small polygons formed is $C_{n-1}^{2}+C_{n}^{4}$.
[Solution 2] When $n=k$, let the total number of small polygons formed by the diagonals of the convex $k$-sided polygon be $m_{k}$. When $n=k+1$, we consider the vertex $A_{k+1}$ as the newly added one, and examine how many more small polygons are added due to its addition.

First, the vertex $A_{k+1}$ draws $k-2$ diagonals, which divide the newly added $\triangle A_{k+1} A_{1} A_{k}$ into $k-1$ small triangles, all of which are newly added. Second, the diagonal $A_{k+1} A_{j}(j=2, \cdots, k-1)$ divides the remaining vertices of the convex $k+1$-sided polygon into two groups: $\left\{A_{1}, A_{2}, \cdots, A_{j-1}\right\}$ and $\left\{A_{j+1}, A_{j+2}, \cdots, A_{k}\right\}$. A diagonal connecting one point from each group intersects with $A_{k+1} A_{j}$. Since no three diagonals intersect at the same point, the number of newly added small polygons (excluding those already counted) is equal to the number of new intersection points of the diagonals. The total number of these new intersection points is
$$
\begin{aligned}
\sum_{m-1}^{k-2} m(k-1-m) & =\frac{1}{2}(k-1)^{2}(k-2)-\frac{1}{6}(k-1)(k-2)(2 k-3) \\
& =C_{k}^{3} .
\end{aligned}
$$

Thus, we obtain the recurrence relation
$$
m_{k+1}=m_{k}+(k-1)+C_{k}^{3} .
$$

From this recurrence relation, we get
$$
\begin{aligned}
m_{n} & =m_{3}+2+3+\cdots+(n-2)+C_{3}^{3}+C_{4}^{3}+\cdots+C_{n-1}^{3} \\
& =C_{n-1}^{2}+C_{4}^{4}+C_{4}^{3}+C_{5}^{3}+\cdots+C_{n-1}^{3} \\
& =C_{n-1}^{2}+C_{n}^{4} .
\end{aligned}
$$
[Solution 3] The total number of diagonals in a convex $n$-sided polygon is
$$
C_{n}^{2}-n=\frac{1}{2} n(n-1)-n=\frac{1}{2} n(n-3)=C_{n-1}^{2}-1 .
$$

Let $m=C_{n-1}^{2}-1$ and order these $m$ diagonals as $l_{1}, l_{2}, \cdots, l_{m}$.
First, remove $l_{1}$, and the total number of small polygons will decrease by $a_{1}+1$, where $a_{1}$ is the number of intersection points of $l_{1}$ with other diagonals (vertices of polygons are not counted). After removing $l_{1}, l_{2}$, $\cdots, l_{j-1}$, remove the diagonal $l_{j}$, and the total number of small polygons will decrease by $a_{j}+1$, where $a_{j}$ is the number of intersection points of $l_{j}$ with the remaining diagonals. When all diagonals are removed, only the original $n$-sided polygon remains. Therefore, the total number of small polygons formed is
$$
\begin{aligned}
\sum_{j=1}^{m \prime}\left(a_{j}+1\right)+1 & =\sum_{j=1}^{m} a_{j}+m+1 \\
& =C_{n}^{4}+\left(C_{n-1}^{2}-1\right)+1=C_{n}^{4}+C_{n-1}^{2} .
\end{aligned}
$$