A7. In the figure, you see two circles and two lines that form nine intersection points. Jaap wants to color exactly four of these nine intersection points red, such that no three red points lie on the same circle or on the same line.

How many possible colorings can Jaap make?
A) 6
B) 12
C) 18
D) 24
E) 36

![](https://cdn.mathpix.com/cropped/2024_04_17_998065f2d90ec501d25cg-2.jpg?height=425&width=414&top_left_y=1047&top_left_x=1415)
A7. C) 18 The middle point can never be colored red, because then there would always be either three red points on the horizontal line or three red points on the vertical line. On both the horizontal and the vertical line, two points must be colored red. There are $\frac{4 \times 3}{2}=6$ ways to color two points red on the horizontal line. In two of the cases, two points are colored red that lie on the same circle. Then there are still two possible points left on the vertical line that can be colored red, and this can only be done in one way. In the other four cases, the two red-colored points lie on different circles, and on the vertical line, one point on the small circle and one point on the large circle must be colored red; this can be done in $2 \times 2=4$ ways. In total, there are thus $2 \times 1+4 \times 4=18$ ways for Jaap to color four points red.