Example 3-19 As shown in Figure 3-9, find the number of paths from point $(0,0)$ to point $(10,5)$ that do not pass through $AB, CD, EF, GH$, given $A(2,2), B(3,2), C(4,2), D(5,2), E(6,2), F(6,3), G(7,2), H(7,3)$. The set of all paths from $(0,0)$ to $(10,5)$ is $S$, and $|S|=\binom{15}{5}=3003$.
Let
$A_{1}:$ the path from $(0,0)$ to $(10,5)$ passes through $A B$,
$A_{2}$ : the path from $(0,0)$ to $(10,5)$ passes through $C D$,
$A_{3}:$ the path from $(0,0)$ to $(10,5)$ passes through $E F$,
$A_{4}:$ the path from $(0,0)$ to $(10,5)$ passes through $G H$.
The number of paths from $(0,0)$ to $(10,5)$ passing through $A B$, according to the multiplication rule, is: the number of paths from $(0,0)$ to $(2,2)$ multiplied by the number of paths from $(3,2)$ to $(10,5)$, so we have
$$
\begin{aligned}
\left|A_{1}\right| & =\binom{4}{2}\binom{(10-3)+(5-2)}{5-2} \\
& =\binom{4}{2}\binom{10}{3}=6 \times 120=720
\end{aligned}
$$

Similarly,
$$
\begin{array}{c}
\left|A_{2}\right|=\binom{6}{2}\binom{8}{3}=15 \times 56=840 \\
\left|A_{3}\right|=\binom{8}{2}\binom{6}{2}=28 \times 15=420 \\
\left|A_{4}\right|=\binom{9}{2}\binom{5}{2}=36 \times 10=360 \\
\left|A_{1} \cap A_{2}\right|=\binom{4}{2}\binom{8}{3}=6 \times 56=336 \\
\left|A_{1} \cap A_{3}\right|=\binom{4}{2}\binom{6}{2}=6 \times 15=90 \\
\left|A_{1} \cap A_{4}\right|=\binom{4}{2}\binom{5}{2}=6 \times 10=60 \\
\left|A_{2} \cap A_{3}\right|=\binom{6}{2}\binom{6}{2}=15 \times 15=225 \\
\left|A_{2} \cap A_{4}\right|=\binom{6}{2}\binom{5}{2}=15 \times 10=150 \\
\left|A_{3} \cap A_{4}\right|=0 \\
\left|A_{1} \cap A_{2} \cap A_{3}\right|=\binom{4}{2}\binom{6}{2}=6 \times 15=90 \\
\left|A_{1} \cap A_{2} \cap A_{1}\right|=\binom{4}{2}\binom{5}{2}=6 \times 10=60 \\
\left|A_{2} \cap A_{3} \cap A_{4}\right|=0 \\
\left|A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\right|=0
\end{array}
$$

Therefore,
$$
\begin{aligned}
\left|\bar{A}_{1} \cap \bar{A}_{2} \cap \bar{A}_{3} \cap \bar{A}_{4}\right|= & 3003-(720+840+420+360)+(336+90 \\
& +60+225+150)-(90+60) \\
= & 3003-2340+861-150=2374
\end{aligned}
$$

Note: There was a discrepancy in the calculation of $\left|A_{2} \cap A_{3}\right|$ in the original text, which should be 225 instead of 900. The final result has been adjusted accordingly.