Example 2.2.7 The ticket price for a concert is 50 yuan per ticket. Among the customers queuing to buy tickets, there are $m$ customers holding 50-yuan bills and $n$ customers holding 100-yuan bills. The ticket booth has no 50-yuan change prepared. 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 $m \geqslant n$.
Analysis: Use $m$ +1s and $n$ -1s to form a sequence of $m+n$ terms: $a_{1}, a_{2}, \cdots, a_{n+m}$ to represent the queue status, where $a_{i}=+1$ indicates the $i$-th customer holds a 50 yuan bill, and $a_{i}=-1$ indicates the $i$-th customer holds a 100 yuan bill. The queue status where no change is unavailable corresponds to the sequence satisfying:
$$
a_{1}+a_{2}+\cdots+a_{k} \geqslant 0 \quad(k=1,2, \cdots, n+m).
$$

Such sequences are called acceptable sequences, and the problem is transformed into finding the number of acceptable sequences composed of $m$ +1s and $n$ -1s.

Solution: Let $f(m, n)$ be the number of acceptable sequences composed of $m$ +1s and $n$ -1s, and let $g(m, n)$ be the number of unacceptable sequences composed of $m$ +1s and $n$ -1s, then
$$
f(m, n)+g(m, n)=\mathrm{C}_{m+n}^{n}.
$$

For an unacceptable sequence $a_{1}, a_{2}, \cdots, a_{n+m}$, there must exist a smallest $k$ such that
$$
a_{1}+a_{2}+\cdots+a_{k} < 0.
$$
By changing the sign of $a_{i}^{\prime}(i=1,2, \cdots, k)$ and keeping the rest unchanged, we obtain an unacceptable sequence composed of $m$ +1s and $n$ -1s. Therefore, the unacceptable sequences composed of $m$ +1s and $n$ -1s correspond one-to-one with the sequences composed of $m+1$ +1s and $n-1$ -1s, so $g(m, n)=\mathrm{C}_{m+n}^{n-1}$,
$$
\begin{aligned}
f(m, n) & =\mathbf{C}_{m+n}^{n}-\mathbf{C}_{m+n}^{n-1}=\frac{(m+n)!}{n!m!}-\frac{(m+n)!}{(n-1)!(m+1)!} \\
& =\frac{(m+n)!}{(n-1)!m!}\left[\frac{1}{n}-\frac{1}{m+1}\right] \\
& =\frac{(m+n)!}{(n-1)!m!} \frac{m-n+1}{n(m+1)} \\
& =\frac{(m-n+1)}{m+1} \mathbf{C}_{m+n}^{n}.
\end{aligned}
$$

If the $m+n$ customers are "indistinguishable", then the number of queue arrangements where no change is unavailable is $\frac{m-n+1}{m+1} \mathrm{C}_{m+n}^{\prime \prime}$.

If the customers are considered "distinguishable", then the answer is $\frac{m-n+1}{m+1} \cdot m!n!\mathrm{C}_{m+n}^{n}=$ $\frac{m-n+1}{m+1}(m+n)!$.