## Task B-4.4.

Points $A, B$, and $C$ are the vertices of a triangle. On side $\overline{A B}$, 6 points are marked, on side $\overline{B C}$, 7 points are marked, and on side $\overline{C A}$, 8 points are marked. The vertices of the triangle are not among the marked points. How many different quadrilaterals can we determine whose vertices are the marked points?
## Solution.

To obtain a quadrilateral whose vertices are the marked points, we have two possibilities:

a) We choose two vertices of the quadrilateral on one side of the triangle, and two on one of the remaining two sides.

There are $\binom{6}{2}\binom{7}{2}+\binom{7}{2}\binom{8}{2}+\binom{6}{2}\binom{8}{2}=1323$ such quadrilaterals.

b) We choose two vertices of the quadrilateral on one side of the triangle, and one vertex on each of the remaining two sides of the triangle.

There are $\binom{6}{2} \cdot 7 \cdot 8+\binom{7}{2} \cdot 6 \cdot 8+\binom{8}{2} \cdot 6 \cdot 7=3024$ such quadrilaterals.

The total number of desired quadrilaterals is $1323+3024=4347$.

Note: The student receives:

- 1 point for dividing into two cases
- 1 point for knowing how to calculate the number of ways to choose two vertices on one side
- 1 point for correctly setting up the products under a)
- 1 point for correctly setting up the products under b)
- 1 point for correctly summing all the products under a) and b)
- 1 point for the correct final result