Example 1 Given a non-negative integer $n$. Choose $(n+1)^{2}$ sets $S_{i, j} \subseteq\{1,2, \cdots, 2 n\}(i, j=0,1, \cdots, n)$ satisfying the following two conditions:
(1) For any $i, j \in\{0,1, \cdots, n\}$, the set $S_{i, j}$ has $i+j$ elements;
(2) For any $0 \leqslant i \leqslant k \leqslant n, 0 \leqslant j \leqslant l \leqslant$ $n$, we have $S_{i, j} \subseteq S_{k, l}$.
Find the number of all different ways to choose. ${ }^{[1]}$
(2019, USA Mathematical Olympiad)
【Analysis】This problem can be expressed in the following form:
In an $(n+1) \times(n+1)$ grid, the bottom-left corner has no number, and the top-right corner has all $2 n$ numbers. Each cell contains all the numbers of the cell directly below it plus one more number, and all the numbers of the cell directly to the left of it plus one more number.
Now we calculate the number of different ways to fill the grid.
Consider the shortest path from the bottom-left corner to the top-right corner, taking the leftmost column and the topmost row.

These $2 n+1$ cells each contain all the numbers of the previous cell plus one more number.
The first cell has no number. Consider the remaining $2 n$ cells.
If we remove all the numbers from the previous cell, the remaining numbers are exactly a permutation of $1,2, \cdots, 2 n$; conversely, if we are given a permutation of $1,2, \cdots, 2 n$, and then fill in all the numbers from the previous cell into the next cell in sequence, we get a valid way to fill the numbers.

This establishes a one-to-one correspondence, so there are $(2 n)!$ ways to fill these $2 n+1$ cells.

Take any one of these fillings, and consider the remaining cells, which are $n^{2}$ in number.

If the left and top adjacent cells of a cell are already determined, since the top adjacent cell has exactly two more numbers than the left adjacent cell, the additional number in this cell compared to the left adjacent cell must be one of these two numbers, and both cases satisfy the requirements.
Therefore, this cell has 2 different ways to be filled.
By the multiplication principle, the total number of different ways to select is $(2 n)!\cdot 2^{n^{2}}$.