Example 5 At different times of the day, the manager hands documents to the secretary for printing, each time placing the document on top of the stack of documents the secretary needs to print. The secretary takes one document from the top to print whenever there is time. If there are $n$ documents, and the manager hands them over in the order $1,2, \cdots, n$, how many possible orders are there for the secretary to finish printing these $n$ documents?
【Analysis】For problems involving counting the return to state 0 after an equal number of "release" and "take out" operations, we can attempt to reduce the problem to the "line method."

Starting from the origin $O$, when a manager delivers a file, draw a line segment with a slope of 1 and a length of $\sqrt{2}$. When a secretary finishes printing a file, draw a line segment with a slope of -1 and a length of $\sqrt{2}$. This way, each order in which the secretary prints the $n$ files corresponds to a line connecting point $O$ and point $M(2n, 0)$ using these two types of line segments, as shown in Figure 1.

Clearly, the directed line segments $OA$ and $BM$ are fixed, meaning points $A(1,1)$ and $B(2n-1,1)$ are mandatory points. Thus, the number of such lines is determined by the ways to connect points $A$ and $B$ using the two types of line segments, totaling $\mathrm{C}_{2n-2}^{n-1}$ lines. This is a total control data, where the lines that cross the $x$-axis (intersecting the line $y=-1$, as shown in Figure 2) are not valid, and thus the number of lines should be reduced.

For such lines, there always exists a common point $K$ with the line $y=-1$ on the far left. Reflect the part of the line between points $A$ and $K$ about the line $y=-1$, resulting in a line segment connecting point $A'(1,-3)$ and point $B(2n-1,1)$. Among these, the upward-sloping line segments are $n+1$ in number, totaling $\mathrm{C}_{2n-2}^{n+1}$ lines.
Therefore, the number of valid lines is
$$
\begin{array}{l}
\mathrm{C}_{2n-2}^{n-1}-\mathrm{C}_{2n-2}^{n+1} \\
=\frac{(2n-2)!}{(n-1)!(n-1)!}-\frac{(2n-2)!}{(n+1)!(n-3)!} \\
=\frac{2}{n+1} \mathrm{C}_{2n-1}^{n} .
\end{array}
$$

Thus, the number of different orders is $\frac{2}{n+1} \mathrm{C}_{2n-1}^{n}$ kinds.