## Task 3 - 300823

Someone wants to draw a number $n$ of points in a plane. They should be chosen such that no three of these points lie on a common line. Afterwards, he wants to find triangles whose all three vertices belong to the drawn $n$ points.

Determine the smallest number $n$ of such points for which it is possible to find 120 different such triangles!
If one has $n$ points as indicated, one can visit any of the remaining $n-1$ points for each of these $n$ points, thus obtaining exactly $n \cdot(n-1)$ ordered pairs of points.

Similarly, one can visit any of the $n-2$ points not included in each of these pairs, thus obtaining exactly $n \cdot(n-1) \cdot(n-2)$ ordered combinations (triples) of three of the $n$ points.

With these combinations, the vertex set of each triangle to be found is captured, and each set is captured exactly 6 times.

For if $A, B, C$ are the vertices of such a triangle, they are captured exactly with the 6 ordered combinations $A B C, A C B, B A C, B C A, C A B, C B A$. Therefore, a total of $\frac{1}{6} \cdot n \cdot(n-1) \cdot (n-2)$ triangles can be found, all of whose vertices belong to the $n$ points.

For $n=10$, this number is 120, for $n<10$ it is less than 120. The smallest number $n$ for which it is possible to find 120 such triangles is therefore $n=10$.