Zhendarov R.G.

Consider an $8 \times 8$ board, the cells of which are not yet colored. In how many ways can the board be colored in black and white so that there are 31 black cells and no two black cells share a side? (Two colorings are considered different if there is a cell that is white in one of these colorings and black in the other.)
Let's divide the board into 16 $2 \times 2$ squares. It is clear that one of the squares is special, containing only one black cell, while the other fifteen squares each contain two black cells, located in opposite corners. The coloring of a non-special square uniquely extends to an adjacent (by side) non-special square, and thus to all non-special squares. Without the special square, we get a chessboard (or its opposite in coloring) with a hole. If the hole is in the top-left (see the figure) or bottom-right corner of the chessboard, then one cell can be colored black in three ways, and in all other cases, in only two ways. Therefore, from a chessboard, we can obtain $16 \cdot 2 + 2 = 34$ colorings, and from its opposite, the same number.

![](https://cdn.mathpix.com/cropped/2024_05_06_73983c7cc26f0024f238g-28.jpg?height=331&width=338&top_left_y=1255&top_left_x=860)

## Answer

68 ways.