【Example 3】Let $n$ be a positive integer. A particle starts at the origin $A(0,0)$ and ends at $B(n, n)$, making non-decreasing moves. The entire path lies below the "diagonal" and touches it at most. Try to find the number of such paths.
The solution is located below the "diagonal," i.e., below the line segment $AB$. It touches $AB$ at most, meaning it only passes through the points in the set $\{(i, j): 1 \leqslant j \leqslant i \leqslant n\}$.

Obviously, the particle needs to move $2n$ steps to move from $A$ to $B$, with $n$ steps to the right and $n$ steps upwards. Therefore, there are $C_{2 n}^{n}$ different non-decreasing paths.

We now calculate the number of non-decreasing paths that satisfy the condition. For each integer $k$ $(0 \leqslant k \leqslant 2 n)$, let $y_{k}$ represent the difference between the number of steps moved to the right and the number of steps moved upwards in the $k$ steps taken. Thus, for each path, we have $y_{0}=0$ and $y_{2 n}=0$. We take another Cartesian coordinate system and sequentially mark the points $(0,0)$, $\left(1, y_{1}\right),\left(2, y_{2}\right), \cdots,(2 n, 0)$, and connect them to form a broken line. This results in a path starting at $(0,0)$ and ending at $(2 n, 0)$, with the entire broken line located above the $x$-axis, intersecting the $x$-axis at most. According to the result of Example 1, the number of such paths is $C_{2 n}^{n} /(n+1)$.
The number $C_{2 n}^{n} /(n+1)$ is typically referred to as the Catalan number, denoted as $C_{n+1}$.