Question 1 At different times during a day, the manager hands documents to the secretary for printing, each time placing a document on top of the stack of documents the secretary is to print. The secretary, when available, takes the document from the top of the stack to print, taking one at a 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 print these $n$ documents?
Solve: Record the secretary's printing order using a grid paper. The method is as follows.

Rule: If the manager places another document on top of the stack of documents to be printed, draw a line segment with a slope of 1 to fill one grid, and call this operation $A$; if the secretary takes the top document from the stack to print, draw a line segment with a slope of -1 to fill one grid, and call this operation $B$.

Next, perform operation $A$ or operation $B$ in the order of time, and the start point of each line segment drawn should connect to the end point of the previous line segment. Thus, you can start drawing from a point $(-n, 0)$ in a selected Cartesian coordinate system. After a day's work, the graph drawn will be a broken line starting from $(-n, 0)$ and ending at $(n, 0)$, which lies on or above the $x$-axis, as shown in Figure 1.
To find the number of such broken lines:
First, out of the $2n$ operations, choose $n$ to perform operation $A$, which determines a broken line from $(-n, 0)$ to $(n, 0)$ (the graph can be below the $x$-axis), with a total of $\mathrm{C}_{2 n}^{n}$ lines.

Second, calculate the number of broken lines from $(-n, 0)$ to $(n, 0)$ that have a part below the $x$-axis. In such a broken line, since it has a part below the $x$-axis, the line intersects with the line $y=-1$. Let the intersection point with the smallest $x$-coordinate be $M(m,-1)$, as shown in Figure 2.

Let the set of all broken lines that have a part below the $x$-axis be $S$.

For each element in $S$, perform a mapping: reflect the part of the broken line to the right of the line $x=m$ about the line $y=-1$, and let the set of the resulting broken lines be $T$.
Then, $|S|=|T|$.
For each broken line in $S$, from $(-n, 0)$ to $(m,-1)$, it performs $\frac{m+n-1}{2}$ operations $A$ and $\frac{m+n+1}{2}$ operations $B$; from $(m,-1)$ to $(n, 0)$, it performs $\frac{n-m+1}{2}$ operations $A$ and $\frac{n-m-1}{2}$ operations $B$.

After the mapping, the image from $(-n, 0)$ to $(m,-1)$ performs $\frac{m+n-1}{2}$ operations $A$ and $\frac{m+n+1}{2}$ operations $B$; from $(m,-1)$ to $(n,-2)$, it performs $\frac{n-m-1}{2}$ operations $A$ and $\frac{n-m+1}{2}$ operations $B$.

Thus, from $(-n, 0)$ to $(n,-2)$, it performs $n-1$ operations $A$ and $n+1$ operations $B$.

Out of the $2n$ operations, choosing $n-1$ to perform operation $A$ determines a broken line in $T$.
Therefore, $|T|=\mathrm{C}_{2 n}^{n-1}$.
Thus, $|S|=|T|=\mathrm{C}_{2 n}^{n-1}$.
The number of broken lines from $(-n, 0)$ to $(n, 0)$ that lie entirely above the $x$-axis is $\mathrm{C}_{2 n}^{n}-\mathrm{C}_{2 n}^{n-1}$.