Example 2 There are $2 n$ people queuing to buy tickets at the theater box office. Each ticket costs 5 yuan, and among them, $n$ people each hold a 5-yuan note, while the other $n$ people each hold a 10-yuan note. The ticket booth has no change. How many ways are there to queue so that everyone can buy their tickets smoothly without encountering any difficulty in giving change?
In the Cartesian coordinate system, let $O(0,0)$ and point $B(2 n, 0)$ represent the "start of ticket sales" and "end of ticket sales," respectively. Buying a ticket with a 5 yuan note is represented by a segment of a broken line with a slope of 1, and buying a ticket with a 10 yuan note is represented by a segment of a broken line with a slope of -1. Thus, for the ticket sales to proceed smoothly, the broken line must remain above the $x$-axis (it can touch but not cross the $x$-axis). From Theorem 1, the number of broken lines from $(0,0)$ to $(2 n, 0)$ is $C_{2 n}^{n}$. To find the number of broken lines that meet the condition, we need to subtract the number of broken lines that cross the $x$-axis. Clearly, each broken line that crosses the $x$-axis must intersect the line $y=-1$ at some point. As shown in Figure 1-7, let the first intersection point from left to right be $A$. By reflecting the segment of the broken line $l$ from $O$ to $A$ across the line $y=-1$ and leaving the rest unchanged, we obtain a broken line $l^{\prime}$ from $O^{\prime}(0,-2)$ through $A$ to $B(2 n, 0)$. Thus, we let $l$ correspond to $l^{\prime}$, which is clearly a one-to-one correspondence. Therefore, the number of broken lines that cross the $x$-axis is equal to the number of broken lines from $O^{\prime}(0,-2)$ to $B(2 n, 0)$, which is $C_{2 n}^{n+1}$. Hence, the number of broken lines that remain above the $x$-axis (or touch it) is $C_{2 n}^{n}-C_{2 n}^{n+1} = \frac{(2 n)!}{n!(n+1)!}$. Therefore, the number of valid queuing methods is $\frac{(2 n)!}{n!(n+1)!}$.