Example 1-44 To go from $(0,0)$ to $(m, n)$, where $m<n$. It is required that all points $(a, b)$ on the path satisfy $a<b$. How many different paths are there?
The number of paths from $(0,0)$ to $(m, n)$ was given in Section 1.9 as $C(m+n, m)$. Now, add the condition that the path must not touch the points on the line $y=x$, and certainly not cross it. This means that the first step from $(0,0)$ must be to $(0,1)$, and not to $(1,0)$, as shown in Figure 1-19.

The problem can also be stated as finding the number of paths from $(0,1)$ to $(m, n)$, where each point $(a, b)$ on the path satisfies $a < b$.

Since $m < n$, it is clear that every path from $(1,0)$ to $(m, n)$ must cross the points on the line $y=x$. We will establish a one-to-one correspondence between each path from $(1,0)$ to $(m, n)$ (Figure 1-20) and each path from $(0,1)$ to $(m, n)$ that passes through a point on the line $y=x$.

From Figure 1-20, if a path from $(1,0)$ to $(m, n)$ intersects the line $y=x$ at points $P_{1}, P_{2}, \cdots, P_{k}$ from left to right, and $P_{k}$ is the last point on the line $y=x$, we can draw a path (solid line) from $(0,1)$ to $P_{k}$ that is symmetric to the path (dashed line) from $(1,0)$ to $P_{k}$ with respect to the line $y=x$. Thus, for each path from $(1,0)$ to $(m, n)$, there is a corresponding path from $(0,1)$ to $(m, n)$ that passes through a point on the line $y=x$. Conversely, for each path from $(0,1)$ to $(m, n)$ that passes through a point on the line $y=x$, there is a corresponding path from $(1,0)$ to $(m, n)$.

Therefore, the number of paths we seek is: $N=$ the number of paths from $(0,1)$ to $(m, n)$ - the number of paths from $(1,0)$ to $(m, n)$.
$$
\begin{aligned}
N & =\binom{m+n-1}{m}-\binom{m+n-1}{m-1} \\
& =(m+n-1)!\left[\frac{1}{m!(n-1)!}-\frac{1}{(m-1)!n!}\right] \\
& =\frac{(m+n-1)!}{m!n!}(n-m)
\end{aligned}
$$