## Task 6 - 300836

In space, twelve points are arranged such that no four of these points lie in a common plane.

Determine the number of all those different tetrahedra whose four vertices belong to the twelve given points!

Hint: Each tetrahedron is uniquely determined by the set of its four vertices (which do not lie in a common plane); there are no requirements regarding the order or the mutual distances of the vertices.
For each of the twelve points, one can pair it with any of the remaining eleven points, thus obtaining exactly $12 \cdot 11$ ordered pairs. Similarly, one can obtain exactly $12 \cdot 11 \cdot 10$ and $12 \cdot 11 \cdot 10 \cdot 9$ ordered combinations of three and four of the twelve points, respectively.

With these combinations of four points, the set of vertices of each of the tetrahedra to be considered is captured, and each set is captured 24 times. If $A, B, C, D$ are the vertices of one of the tetrahedra, then all ordered combinations of these points are captured. The number of these can be determined as follows:

Any of the four points $A, B, C, D$ can be placed in the first position of a combination, and for each of these possibilities, there are three points left to choose from for the second position. For each of the resulting $4 \cdot 3$ possibilities, there are two choices left for the third position, and in each of the $4 \cdot 3 \cdot 2 = 24$ resulting possibilities, the fourth position is also determined.

Thus, the number of tetrahedra sought is

$$
\frac{12 \cdot 11 \cdot 10 \cdot 9}{24} = 11 \cdot 5 \cdot 9 = 495
$$

Solutions of the third round 1990 taken from [5]

### 5.33 XXXI. Olympiad 1991

### 5.33.1 First Round 1991, Class 8