【Example 5】As shown in Figure 21, let $n<m$ be positive integers, a particle moves from the origin $A(0,0)$ to the point $C(m, n)$ in a non-decreasing manner, and let $B(n, n)$.
(i) Find the number of paths that lie entirely below $A B$, and touch it at most once;
(ii) Find the number of paths that have no common points with $A B$.


Method 1: Still convert according to the previous method, obtaining a path starting at $(0,0)$ and ending at $(m+n, m-n)$, and now the path, except for the starting point on the $x$-axis, has no other common points with the $x$-axis (all points are above the line $y=1$, and at most tangent to $y=1$).
Such a path must pass through the point $(1,1)$. We translate the origin of the coordinate system to $(1,1)$, so the path becomes one starting at $(0,0)$ and ending at $(m+n-1, m-n-1)$, located above the $x$-axis and at most tangent to the $x$-axis. Therefore, by Example 1, the number of such paths is $\left[(m-n) C_{m+n-1}^{n}\right] / m$.
Method 2: First, find the number of paths $l$ that intersect $AB$. Clearly, any path that first reaches the point $(0,1)$ and finally reaches point $C$ must intersect $AB$; and for each path that first reaches the point $(1,0)$ and intersects $AB$, if the first intersection point with $AB$ is denoted as $T$, and the part of the broken line between $T$ and $A(0,0)$ is reflected about the line $y=x$, it results in a path that first reaches the point $(0,1)$ and then reaches point $C$; conversely, the same applies. Therefore, the number of paths $l$ that intersect $AB$ is equal to twice the number of paths $l_{1}$ that first reach the point $(0,1)$ (and then reach point $C$). And
$$
\left.\frac{l_{1}}{C_{m+n}^{m}}=P \text { (first step up }\right)=\frac{n}{m+n}
$$

Thus, $l=2 l_{1}=2 n C_{m+n}^{m} /(m+n)=2 C_{m+n-1}^{n}$, so the number of paths that meet the requirements is
$$
C_{m+n}^{m}-l=\frac{m-n}{m+n} C_{m+n}^{m}=\frac{m-n}{m+n} \frac{(m+n)!}{m!n!}=\frac{m-n}{m} C_{m+n-1}^{n}
$$