$5 \cdot 67$ At the theater ticket office, $2n$ tickets are sold, each ticket costing 50 cents, and each person is limited to buying one ticket. Initially, the ticket office has no money, and among the first $2n$ people in the queue, half have 50-cent coins, and the other half only have one-dollar bills. To ensure the ticket office does not encounter any difficulty in giving change, how many different ways can these $2n$ ticket buyers be arranged in the queue?
[Solution] In the Cartesian coordinate system $x o y$, let the origin $O$ represent the ticket office, and the ticket buyers are represented by points with horizontal coordinates $1, 2, \cdots, 2n$.

Let each ticket buyer with one yuan correspond to the number 1, and each ticket buyer with fifty cents correspond to the number -1.

Add the number corresponding to each ticket buyer and all the ticket buyers in front of him, and use the sum as its vertical coordinate, representing it as a point on the plane, then connect adjacent points with line segments, obtaining a broken line (as shown by the solid line in the figure).

Since there are $n$ people with one yuan and $n$ people with fifty cents, this broken line starts from $(0,0)$ and ends at $(2n, 0)$. It has $n$ segments rising and $n$ segments falling, so there are $C_{2n}^{n}$ such broken lines.

It is not difficult to see that a broken line represents a queuing method, and vice versa. Therefore, there are a total of $C_{2n}^{n}$ different queuing methods.

The queuing methods that cause change difficulties correspond to broken lines that touch the line $y=1$. To calculate the number of broken lines that touch the line $y=1$, for all such broken lines, starting from the first point of contact, flip the subsequent part of the broken line with the line $y=1$ as the axis (as shown by the dashed line in the figure). This way, a broken line from $(0,0)$ to $(2n, 2)$ is obtained. It is not difficult to see that each broken line that touches the line $y=1$ corresponds to a broken line from $(0,0)$ to $(2n, 2)$, and vice versa. However, a broken line from $(0,0)$ to $(2n, 2)$ has $(n+1)$ segments rising and $(n-1)$ segments falling, so there are $C_{2n}^{n-1}$ such broken lines, i.e., from $(0,0)$ to $(2n, 0)$, and touching the line $y=1$, there are $C_{2n}^{n-1}$ broken lines, i.e., there are $C_{2n}^{n-1}$ queuing methods that cause change difficulties.
Therefore, the number of queuing methods that do not cause change difficulties is
$$
C_{2n}^{n} - C_{2n}^{n-1} = \frac{1}{n} C_{2n}^{n-1} \text{ methods. }
$$