We paint the faces of a cube with 6 different colors. The faces are labeled with numbers from 1 to 6, such that the pairs of opposite faces are labeled as follows: 1 and 6, 3 and 4, 2 and 5. How many different cubes can we create this way?
I. Solution: We paint one of the faces with a certain color. For the opposite face, we can choose from 5 colors. When painting any of the remaining four faces, which form a ring, with a color not yet used, we essentially arrive at the same cube. The remaining 3 faces can be painted with the remaining colors in $3!=6$ ways. Therefore, in terms of coloring, there are $5 \cdot 6=30$ cases.

On the painted cube, 3 pairs of faces can be assigned 3 pairs of numbers $(1,6 ; 2,5 ; 3,4)$ in 3! ways. However, the pairs themselves can also be swapped, which in each case results in 2 additional cases. Thus, the numbers can be placed on the colored cube in 3! $2^{3}=6 \cdot 8=48$ ways.

Accordingly, the total number of different cubes that can be made this way is $30 \cdot 48=1440$.

István Molnár (Debrecen, Ref. g. III. o. t.)

II. Solution: First, let's examine how many ways a cube can be numbered without coloring. Since according to our condition, the 1 face must always be adjacent to the 2 face, the cube can always be placed on the table so that the 1 face is on top and the 2 face is facing us. This determines the positions of the 6 and 5 faces as well. The 3 can now be on the left or the right. Therefore, the numbering of the uncolored cube can only occur in 2 different ways.

After the numbering is done, the 6 different colors can clearly be placed on the 6 numbered faces in as many ways as the permutations of the 6 colors. The number of these is $6!=720$, and thus the total number of different cubes is $2 \cdot 720=$ 1440.

András Bárdos (Bp., II., Rákóczi g. III. o. t.)