The seventy-sixth problem: Given a non-negative integer $\mathrm{n}$, select $(\mathrm{n}+1)^{2}$ sets $\mathrm{S}_{\mathrm{i}, \mathrm{j}} \subseteq\{1,2, \ldots, 2 \mathrm{n}\}$ (i, $\mathrm{j} \in$ $\{0,1, \ldots, \mathrm{n}\}$ ), satisfying the following two conditions:
(1) For any $i, j \in\{0,1, \ldots, n\}$, the set $S_{i, j}$ contains exactly $i+j$ elements;
(2) For any $0 \leq \mathrm{i} \leq \mathrm{k} \leq \mathrm{n}, 0 \leq \mathrm{j} \leq \mathrm{t} \leq \mathrm{n}$, we have $\mathrm{S}_{\mathrm{i}, \mathrm{j}} \subseteq \mathrm{S}_{\mathrm{k}, \mathrm{t}}$.
Find the number of all different selection methods.
The 76th question:
Solution: This problem can be stated 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 $2n$ 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 $2n+1$ cells each contain all the numbers of the previous cell plus one more number. The first cell has no number. Considering the remaining $2n$ cells, the one additional number in each of these $2n$ cells forms a permutation of $1, 2, \ldots, 2n$. There are $(2n)!$ such permutations.

Assume that the additional number in each of these $2n$ cells forms the permutation $(1, 2, \ldots, 2n)$. Consider the remaining $n^2$ cells. If the left and top adjacent cells of a cell are already determined, since the top adjacent cell contains exactly two more numbers than the left adjacent cell, the additional number in this cell must be one of these two numbers, and both cases satisfy the condition. Thus, each of the remaining $n^2$ cells has exactly two valid ways to be filled.
In summary, according to the multiplication principle, there are exactly $(2n)! \cdot 2^{n^2}$ ways to satisfy the conditions.