TITLE: How to count the no. of distinct ways in which $1,2,...,6$ can be assigned to $6$ faces of a cube?
QUESTION [6 upvotes]: How to find the number of ways in which six digits $1,2,..,6$ can be assigned to six faces of a cube (without repetition of digits) so that one arrangement cannot be obtained from another by a rotation of the cube?
I tried to find the number of unique $4-$adjacencies of the faces of the cube. I drew a simple undirected graph $A,B,...,F$ having $6$ vertices with each of them having degrees equal to $4$. There were $12$ edges. Considering the choice between the top and the bottom of the cube, I found the result to be $12 \times 2=24.$
Am I correct? Please suggest better approach approach or bijections(if there are any).

REPLY [3 votes]: Using the Polya Enumeration Theorem and cycle indices we can say that the cycle index of the face permutation group $G$ of the cube contains the identity plus some other permutations ($23$ of them) that we don't need to compute exactly because there is a cycle of length at least two in their disjoint cycle decomposition and hence the subsitution with $x_1+x_2+\cdots+x_6$ produces some digit more than once. Thus the only contributing permutation is the identity with a contribution of $$\frac{1}{24} a_1^6.$$
This gives
$$[x_1 x_2\cdots x_6]Z(G)(x_1, x_2, \ldots x_6)
= \frac{1}{24} [x_1 x_2\cdots x_6] (x_1+x_2+\cdots+x_6)^6 = \frac{1}{24}\times 6! = 30.$$