## Task 3 - 300813

In a plane, seven points are given such that no three of them lie on a common line.

Determine the number of all those triangles whose vertices are three of the given points!
If for each of the 7 points one of the other 6 is written down, and for each of the thus written down $7 \cdot 6$ pairs one of the other 5 points, one obtains $7 \cdot 6 \cdot 5$ combinations of 3 points each.

Each such combination gives the vertices of one of the sought triangles. In this way, each of these triangles is captured, and exactly 6 times.

For example, if $A, B, C$ are the vertices of a sought triangle, it will be captured exactly with the combinations $A B C, A C B, B A C, B C A, C A B, C B A$.

The number of all sought triangles is therefore one-sixth of $7 \cdot 6 \cdot 5$, i.e., it is $7 \cdot 5 = 35$.