10.2. In the bottom left corner of a $7 \times 7$ chessboard, there is a king. In one move, he can move one square to the right, or one square up, or one square diagonally to the right and up. In how many different ways can the king travel to the top right corner of the board, if he is forbidden to visit the central cell?
Answer: 5020.

Solution. We will construct a $7 \times 7$ table, in each cell of which we will write the number equal to the number of allowed paths by which the king can reach this cell from the bottom left corner. First, we fill the left column and the bottom row with ones and write 0 in the central cell (as per the condition). Then we fill the second column from the left and the second row from the bottom, and so on, according to the rule: in the current cell, we place the sum of the numbers in the three adjacent cells below, to the left, and diagonally (below and to the left). As a result, we get the table

| 1 | 13 | 85 | 314 | 848 | 2078 | 5020 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 11 | 61 | 168 | 366 | 864 | 2078 |
| 1 | 9 | 41 | 66 | 132 | 366 | 848 |
| 1 | 7 | 25 | 0 | 66 | 168 | 314 |
| 1 | 5 | 13 | 25 | 41 | 61 | 85 |
| 1 | 3 | 5 | 7 | 9 | 11 | 13 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |

The answer is the number in the top right corner, 5020.