## Task $7 / 87$

Determine how many paths of length 7 exist on the edges of a unit cube from a specified corner to the diametrically opposite corner; no edge may be traversed more than once (not even in the opposite direction!), but this restriction does not apply to the corners.
The corners of the cube are labeled $A ; B ; C ; D ; E ; F ; G ; H$ such that $A B C D$ is a cube face and $A G, B H, C E$ and $D F$ are space diagonals.

Starting point is $A$, endpoint is $G$. From $A$ there are 3 (due to symmetry) equivalent paths; we follow the one to $B$ (and multiply the result by 3). From $B$ there are 2 equivalent paths; we go to $C$ (and multiply the result by 2). Here, we can take 2 paths again, but they are not equivalent: $C G$ or $C D$.

1. $C G$ leads to the goal in one path of length 3. For a path of length 7, one must still go around one side face. Since the faces $G F B C$ and $G C D H$ would lead to double traversal of edges, the path can only continue along the third side face. Thus, 2 valid paths are found: $A B C G F E H G$ and $A B C G H E F G$.
2. We check $C D$.

2.1. $A B C D H G$ has only a length of 5; a "detour" over $E F$ leads to the 3rd valid path: $A B C D H E F G$. 2.2. $A B C D A E$ yields two more possibilities: $A B C D A E F G$ and $A B C D A E H G$.

Thus, 5 valid paths are determined via $A B C$. There are therefore $2 \cdot 5=10$ valid paths via $A B$ and thus a total of $3 \cdot 10=30$.