10.5. Find the number of all possible arrangements of chips in some cells of an 8 by 8 chessboard such that the number of chips in each row is different and the number of chips in each column is different.
Answer: $2 \cdot(8!)^{2}$.

Solution. A row or column can contain from 0 to 8 chips, so if the quantities in the rows and columns are different, then the set of values of the number of chips in the rows and the set of values of the number of chips in the columns both represent a sequence of numbers $0,1,2$, $\ldots, 7,8$ with one missing number, and this missing number is the same in both cases and equals the difference of 36 and the number of chips on the board. If this missing number is not 0 or 8, then there will be either an empty row and a full column, or vice versa, on the board at the same time, which is impossible. Therefore, there are two possibilities: there are 36 chips on the board, and the quantities of chips in the rows and columns are $1,2,3, \ldots, 8$, or there are 28 chips on the board, and the quantities of chips in the rows and columns are $0,1,2,3, \ldots, 7$. By simultaneously swapping the occupied cells with empty cells and vice versa, we see that the number of different arrangements of the first and second types are equal.

Let us be given an arbitrary arrangement of the first type, in which there will be a filled row and a filled column - a total of ${ }^{8 \cdot 8}=64$ options for choosing their numbers. We strike them out and get an arrangement of 21 chips on a 7 by 7 board with different numbers of chips in the rows and different numbers of chips in the columns. On the 7 by 7 board, this corresponds to an arrangement with an empty row and an empty column, for which there are $7 \cdot 7=49$ ways to choose. We strike them out again, getting an arrangement of 21 chips on a 6 by 6 board, and so on. Continuing this way, we get $(8!)^2$ options for choosing the numbers of the corresponding rows and columns (alternately full and empty), which completely and uniquely determine the given arrangement of 36 chips on an 8 by 8 board. Adding to them the same number of arrangements of 28 chips on 8 by 8, we get the answer: $2 \cdot(8!)^{2}$.

Grading criteria. It is proven that there can only be 28 or 36 chips on the board: 2 points. The number of arrangements of only 28(36) chips is correctly calculated: 6 points. Any provably incorrect answer: 0 points.