a) Consider the $4 \times 4$ board shown in the following figure. The letters and numbers have been placed to help locate the small squares. For example, the square $A 1$ is in the top-left corner and the $D 4$ is in the bottom-right corner.

![](https://cdn.mathpix.com/cropped/2024_05_01_4a19aea8e08094ae376fg-29.jpg?height=355&width=343&top_left_y=634&top_left_x=931)

Show a way to choose 12 squares such that exactly 3 squares are chosen in each row and each column.

b) Determine the number of different ways to choose 12 squares on the $4 \times 4$ board such that exactly 3 squares are chosen in each row and each column.

c) Consider an $8 \times 8$ board with a similar identification of rows and columns as the previous board, but colored alternately black and white.

![](https://cdn.mathpix.com/cropped/2024_05_01_4a19aea8e08094ae376fg-29.jpg?height=745&width=737&top_left_y=1536&top_left_x=739)

Determine the number of ways to choose 56 squares from the board such that all the black squares are chosen and exactly 7 squares are chosen in each row and each column.
Solution

a) There are several ways to do this (we will see exactly how many in the next item). In the example on the left below, only the squares $A 1, B 2, C 3$ and $D 4$ are not chosen.
![](https://cdn.mathpix.com/cropped/2024_05_01_4a19aea8e08094ae376fg-30.jpg?height=374&width=1170&top_left_y=756&top_left_x=316)

b) We will use the fundamental principle of counting. In the first row, we have 4 ways to select the square that will not be chosen. For each of these ways, we will have 3 ways to select the square in the second row. Then, 2 ways to select in the third row, and only one way to select in the fourth row. With this, we have $4 \cdot 3 \cdot 2 \cdot 1 = 4! = 24$ ways to choose the 12 squares.

c) If we consider that all the black squares are chosen, it remains to choose exactly 3 white squares in each row (line or column). Note that the number of ways to do this is equal to the number of ways to choose 3 squares in each row of the $4 \times 4$ boards, indicated to the right and in the center in the figure above, such that in each row exactly 3 squares are chosen. Since the choices in each board are independent, for each choice of 12 squares from the board in the center and each choice of 12 squares from the board on the right, meeting all the conditions, we have a choice of 24 white squares from the $8 \times 8$ board. Together with the 32 black ones, they form the 56 that must be chosen. Thus, the number of ways to choose the 56 squares satisfying the conditions of the problem is $24 \cdot 24 = 576$.