## Task A-4.5.

On a board of dimensions $5 \times 7$ where all fields are white, it is necessary to color exactly 17 fields black so that the resulting arrangement of black and white fields on the board is centrally symmetric, i.e., does not change by a $180^{\circ}$ rotation around the center of the board. In how many ways can this be done?
## Solution.

If a certain field is black, then its centrally symmetric field (with respect to the center of the board) must also be black, and similarly for white fields. Thus, all fields except the central one can be divided into pairs of centrally symmetric fields of the same color.

Since the total number of black fields (17) is an odd number, the central field must be colored black.

The remaining 16 black fields come in eight pairs of centrally symmetric fields.

![](https://cdn.mathpix.com/cropped/2024_05_30_acf83088931d085246bdg-24.jpg?height=577&width=582&top_left_y=2007&top_left_x=674)

If we divide the board without the central field into two parts that are mutually centrally symmetric (for example, as shown in the figure), then in each of these parts exactly 8 out of 17 fields must be colored black.

Any selection of black fields in one part uniquely determines the black fields in the other part. The solution is the number of ways to choose 8 out of 17 fields, which is $\binom{17}{8}$.

Note: $\binom{17}{8}$ equals 24310, but it is not necessary to calculate this.