Example 1-45 The ticket price for a concert is 50 yuan, and among the customers queuing to buy tickets, there are $n$ customers with 50-yuan bills and $m$ customers with 100-yuan bills. The ticket booth has no 50-yuan change. How many ways are there to queue so that the ticket purchasing process can proceed smoothly without encountering a situation where change cannot be given? Assume each customer buys only one ticket, and $n \geqslant m$.
Assuming that an $m+n$ dimensional 0,1 row vector is used to represent a queue state, let this row vector be
$$
\left(a_{1} a_{2} \cdots a_{m \cdot n}\right)
$$

where $a_{i}=0,1, i=1,2, \cdots, m+n$.
$a_{i}=1$ indicates that the $i$-th customer has a 50 yuan bill; $a_{i}=0$ indicates that the $i$-th customer has a 100 yuan bill.
Thus, a vector with $m$ 0 elements and $n$ 1 elements has a total of $C(m+n, m)$ possibilities.
Each vector can correspond one-to-one with a path from the point $(0,0)$ to the point $(m, n)$, i.e., starting from the point $(0,0)$, when $a_{i}=0$, move one unit in the $x$-axis direction; when $a_{i}=1$, move one unit in the $y$-axis direction.

To ensure that customers can buy tickets smoothly without encountering a situation where 50 yuan bills cannot be found, the points $(x, y)$ on the path must satisfy $x \leqslant y$.

The problem is equivalent to finding the number of paths from the point $(0,0)$ to the point $(m, n)$ that do not cross the line $y=x$ (i.e., finding the points $(x, y)$ on the path such that $y \geqslant x$), where the paths can touch the line $y=x$ but not cross it, or do not touch the line $y=x-1$. According to the previous problem, the number of such paths is the solution to the problem from the point $(0,0)$ to the point $(m, n+1)$, i.e., the number of such paths is
$$
\begin{aligned}
& C(m+n, m)-C(m+n, m-1) \\
= & (m+n)!\left[\frac{1}{m!n!}-\frac{1}{(m-1)!(n+1)!}\right] \\
= & (m+n)!\frac{n-m+1}{m!(n+1)!}
\end{aligned}
$$