A1. We have a field of $4 \times 4$ square cells. Of these 16 cells, we want to color exactly four black. This must be done in such a way that each row and each column gets exactly one black cell, and no two black cells are diagonally (with a corner) adjacent.

In how many ways can we choose the four cells to color?
A) 1
B) 2
C) 3
D) 4
E) This is not possible.

![](https://cdn.mathpix.com/cropped/2024_04_17_9e808ca4ac6657f5611fg-1.jpg?height=300&width=297&top_left_y=1272&top_left_x=1505)
A1. B) 2 Suppose we color field B2 black. Then the 8 surrounding fields may not be colored black anymore: the fields above, below, to the left, and to the right of B2 are in the same column or row as B2, and the other four fields touch B2 diagonally. This leaves only row 4 and column D, and in each, we can only color one field black. In total, no more than 3 fields can be colored black. We conclude that B2 may not be colored black.

In the same way, we can deduce that the fields B3, C2, and C3 may not be colored black either. In row 2, we can therefore only color A2 or D2 black. If we color A2 black, then in row 3 we must color D3 black, because A3 is in the same column as A2. For rows 1 and 4, it is then only possible to color C1 and B4 black. This gives one solution. If we color D2 black instead of A2, we find the solution where D2, A3, B1, and C4 are colored black. In total, there are thus 2 ways to choose the black squares.