Question 184: Fill $n^{2}(n \geq 3)$ integers into an $n \times n$ grid, with one number in each cell, such that the product of the numbers in each row and each column is 30. How many ways are there to do this?
Question 184, Solution: First, consider the case where all integers in the grid are positive integers. In this case, the prime factors of the numbers in each row and each column exactly include one $2$, one $3$, and one $5$. Among these, the $\mathrm{n}$ twos are in different rows and different columns, with $\mathrm{n}$! ways to fill them. Similarly, the threes and fives each have $\mathrm{n}$! ways to fill them. Therefore, in this case, there are $(\mathrm{n}!)^3$ ways to fill the grid.

Furthermore, the numbers in the $(n-1) \times (n-1)$ subgrid in the top-left corner can arbitrarily choose their positive or negative signs. At this point, the numbers in the $n$-th row and the $n$-th column have a unique way to choose their positive or negative signs. Hence, there are $2^{(n-1)^2} \cdot (\mathrm{n}!)^3$ ways to fill the grid that satisfy the conditions.