Problem 2. A square canvas is divided into 100 identical squares, arranged in 10 rows and 10 columns. We have 10 cards, numbered differently with digits from 0 to 9. On the canvas, we need to place two cards, whose sum is 10, in squares located on different rows and columns. Determine the number of possibilities for placing these cards.
Solution and grading: Since the sum of the digits must be equal to 10, there are only 4 pairs of cards that can be placed on the board. ......................................1p

The first card can be placed in any square, so there are 100 options. For the second card, 9 rows and 9 columns remain available, which means 81 options. Therefore, we have $100 \times 81=8100$ ways to place a pair of cards. . . . . . $4 \mathbf{p}$

Since we have 4 pairs, we have a total of $4 \times 8100=32400$ options. ...... $\mathbf{2 p}$