Example 1.11 A and B are playing table tennis, and the final score is $21: 17$. Find the number of scoring scenarios in which A has always been leading B during the game.
Solution: Let the required number be $N$. A scoring scenario uniquely determines a sequence $a_{1} a_{2} \cdots a_{38}$, where
$$
a_{j}=\left\{\begin{array}{l}
1, \text { when the } j \text {th score is for player A; } \\
-1, \text { when the } j \text {th score is for player B. }
\end{array} \quad(1 \leqslant j \leqslant 38)\right.
$$

Let $A_{j}$ represent the point $\left(j, a_{1}+a_{2}+\cdots+a_{j}\right)(j=1,2, \cdots, 38)$. Then, the scoring scenario can be represented by the directed broken line $\overline{A_{1} A_{2} \cdots A_{38}}$. Since player A's score is always ahead of player B, the y-coordinate of $A_{j}$ is greater than zero. Also, the difference in the x-coordinates between $A_{j+1}$ and $A_{j}(j=1,2, \cdots, 37)$ is 1, and the difference in the y-coordinates is
$$
\left(a_{1}+a_{2}+\cdots+a_{j+1}\right)-\left(a_{1}+a_{2}+\cdots+a_{j}\right)=a_{j+1} \text {, }
$$
leading to $A_{38}(38,4)$ and not crossing the x-axis. Therefore, $N$ is equal to the number of $\mathrm{T}$ paths from the point $(1,1)$ to the point $(38,4)$ that do not cross the x-axis. By Theorem 1.10, we have
$$
\begin{aligned}
N & =\frac{(38-1)!}{\left(\frac{38-1}{2}+\frac{4-1}{2}\right)!\left(\frac{38-1}{2}-\frac{4-1}{2}\right)!} \\
& \frac{(38-1)!}{\left(\frac{38-1}{2}+\frac{4+1}{2}\right)!\left(\frac{38-1}{2}-\frac{4+1}{2}\right)!} \\
& =\frac{37!}{20!17!}-\frac{37!}{21!16!}=\frac{4 \times 37!}{21!17!} .
\end{aligned}
$$