We are making colored cubes according to the following specifications.

I. Each face of the cube is divided into two triangles by one of its diagonals, such that each endpoint of the drawn diagonal connects to two other such diagonals.

II. The two triangles on each face must be of different colors.

III. The triangles that meet along an edge must be of the same color.

How many different cubes can we make using 6 colors? (We do not consider two cubes different if they can be positioned so that from any angle, the faces we see are identically colored.)
According to the I. specification, 6 diagonals with 12 endpoints join in threes at 4 vertices (at points $A^{\prime}, B, C^{\prime}$, and $D$ on the diagram), while at the 4 other vertices of the cube, 3 each, according to the III. specification, must be painted in the same color triangles converge. A conical surface formed by a set of three such triangles (as around vertex $B^{\prime}$ in the diagram) touches each of the other three along an edge formed by a face diagonal, so these must be painted in different colors in pairs. Thus, on each cube, 4 colors appear out of the 6 available. This selection can be made in $\binom{6}{4}=15$ ways.

![](https://cdn.mathpix.com/cropped/2024_05_02_9b1d2771bfe42f1ce137g-1.jpg?height=477&width=466&top_left_y=434&top_left_x=819)

Let the 4 colors now be, for example, red, white, green, and black. The first can be placed at will - for example, around $B^{\prime}$ - and the second can also be placed at will, for example, around $A$. However, placing the green at $C$ and $D^{\prime}$ results in different colorings. The black goes to the yet unpainted triangles. Thus, 4 colors can be placed on a cube in 2 ways, and the number of differently colorable cubes is $15 \cdot 2=30$.

Traply Endre (Budapest, Fazekas M. Gyak. Gimn., I. o. t.)