(Concours général 90)

We have $n \geq 4$ colors. How many different tetrahedra can be painted by painting each face with one and only one color?
We denote $1,2, \ldots, n$ as the colors (thus, we have an order on the colors).

We will distinguish the different cases based on the number of colors used.

- If we use 4 different colors, all faces are of different colors. We place the smallest color (the one with the smallest number) at the bottom and the second color facing us. There are then 2 possibilities for choosing the positions of the last 2 colors. Therefore, there are 2 (and only 2) tetrahedra that have the same 4 colors. There are thus $2 \cdot \binom{n}{4}$ tetrahedra with 4 colors.
- If we use 3 colors, there must be one color that repeats, let's denote it as $i$. We then place the two faces of color $i$ at the bottom and the other one facing us, the other two faces can thus be painted in 2 different ways. Therefore, there are 2 $\cdot \binom{n}{3}$ tetrahedra with 3 different colors.
- If we use 2 colors, either one color appears 3 times, or both colors appear 2 times each. If one of the 2 colors appears 3 times, then by placing the color that appears only once at the bottom, we notice that there is only one way to paint the tetrahedron. If each color appears 2 times and we place the smaller color at the bottom and facing us, we notice that the position of the other color is thus fixed. Therefore, there is only one way to paint a tetrahedron with two colors that each repeat twice. There are thus 2 $\cdot \binom{n}{2}$ tetrahedra painted with 2 colors.
- If we use a single color, there is of course only one tetrahedron per color. There are thus $\binom{n}{1}=n$ tetrahedra of a single color.

The number of possible tetrahedra is therefore:

$$
\begin{aligned}
2\left(\binom{n}{4}+\binom{n}{3}\right. & \left.+\binom{n}{2}\right)+n=2\left(\frac{n(n-1)(n-2)(n-3)}{4!}+\frac{n(n-1)(n-2)}{3!}+\frac{n(n-1)}{2}\right)+n \\
& =n\left(\frac{(n-1)(n-2)(n-3)}{12}+\frac{(n-1)(n-2)}{3}+(n-1)+1\right)
\end{aligned}
$$