\section*{Exercise 1 - 341021}

a) How many different distributions of the numbers \(1,2, \ldots, 6\) on the six side faces of a cube are there in total?

b) How many different distributions among these satisfy the additional condition that for each pair of opposite side faces, the numbers on these two faces sum to 7?

Hint: In a) and b), two distributions are considered different if and only if they cannot be transformed into each other by rotating the cube.
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a) We take a cube and label its faces so that each of the numbers from 1 to 6 appears exactly once.

We place the cube so that 1 is on the top face, which leaves 5 possibilities for which number is on the bottom face. Then we rotate the cube around the vertical axis so that the smallest of the four adjacent faces points forward. There are still \(3!=6\) possibilities for how the other 3 faces are labeled. Furthermore, each cube can be placed in a unique way so that 1 is on top and the smallest number of the four adjacent faces points forward. Therefore, there are a total of \(5 \cdot 3!=30\) pairwise different distributions.

b) We take a cube and label its faces so that each of the numbers from 1 to 6 appears exactly once and the numbers on opposite faces add up to 7.

We place the cube so that 1 is on the top face. Since 6 (and not 2) is on the opposite face, we can rotate the cube around the vertical axis so that the face with 2 points forward. Thus, 5 is on the back face. Now, 3 can be on the left or the right face. Therefore, there are a total of 2 possibilities.