Given 19 drawers arranged in a row, place $n$ identical balls into them such that the number of balls in each drawer does not exceed the number of balls in the drawer to its left. Let the number of such arrangements be denoted by $F_{m, n}$.
(1) Find $F_{1, n}$;
(2) If $F_{m, 0}=1$, prove:
$$
F_{m, n}=\left\{\begin{array}{ll}
F_{n, n}, & (m>n \geqslant 1) \\
F_{m-1, n}+F_{m, n-m} . & (1<m \leqslant n)
\end{array}\right.
$$
(3) Calculate $F_{3,8}$.
(1991, Jiangsu Province Mathematical Summer Camp)
Solution: (1) When $m=1$, there is obviously only one way, i.e., $F_{1, n}=1$.
(2) When $m>n \geqslant 1$, for any method that meets the conditions, there will be no balls in all the drawers to the right of the $n$-th drawer, so $F_{m, n}=F_{n, n}$.

When $1$ 0, if we reduce the number of balls in each drawer by 1, we clearly get a method of placing $n-m$ balls into $m$ drawers. Conversely, for any method of placing $n-m$ balls into $m$ drawers, if we increase the number of balls in each drawer by 1, we get a method of placing $n$ balls into $m$ drawers such that $x_{m}>0$. Therefore, $F_{m, n}=F_{m-1, n}+F_{m, n-m}$.
(3) From the result in (2), it is not difficult to deduce that $F_{3,8}=10$.
Example 11 Let $P_{1}$ be a point on the side $AB$ of the equilateral $\triangle ABC$. Draw a perpendicular from $P_{1}$ to the side $BC$, with the foot of the perpendicular being $Q_{1}$. Draw a perpendicular from $Q_{1}$ to the side $CA$, with the foot of the perpendicular being $R_{1}$. Draw a perpendicular from $R_{1}$ to the side $AB$, with the foot of the perpendicular being $P_{2}$, and so on, to get points $Q_{2}, R_{2}, P_{3}, Q_{3}, R_{3}, \cdots$. When $n$ $\rightarrow \infty$, which point does $P_{n}$ approach infinitely?

Solution: As shown in Figure 2, let $B P_{n}=x_{n}, A B=a$, then $B P_{n+1}=x_{n+1}$. Find the relationship between $x_{n}$ and $x_{n+1}$:
$$
\begin{array}{l}
B P_{n}=x_{n} \\
\rightarrow B Q_{n}=\frac{1}{2} x_{n} \\
\rightarrow Q_{n} C=a-\frac{1}{2} x_{n} \\
\rightarrow C R_{n}=\frac{1}{2}\left(a-\frac{1}{2} x_{n}\right) \\
\rightarrow A R_{n}=a-C R_{n}=\frac{1}{2} a+\frac{1}{4} x_{n} \\
\rightarrow A P_{n+1}=\frac{1}{2} A R_{n}=\frac{1}{4} a+\frac{1}{8} x_{n} \\
\rightarrow B P_{n+1}=x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a, \\
\text { hence } x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a .(n \geqslant 1)
\end{array}
$$

Thus, $x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a \cdot(n \geqslant 1)$
It is easy to get $x_{n}=\frac{2}{3} a+\left(x_{1}-\frac{2}{3} a\right)\left(-\frac{1}{8}\right)^{n-1}$
Therefore, $\lim x_{n}=\frac{2}{3} a$, which means that point $P_{n}$ approaches infinitely close to the point that divides $AB$ in the ratio $1:2$.