How to count the no. of distinct ways in which $1,2,...,6$ can be assigned to $6$ faces of a cube?

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).

Solution 1:

The number of permutations of the numbers on the faces is $6!$. The number of ways one of these can be transformed into another by a rotation of the cube is the order of the rotation group of the cube, which is $24$ ($=4!$, since there is exactly one rotation for each permutation of the four pairs of opposite corners). Thus, the number of permutations of the numbers that are inequivalent under rotations is $6!/4!=6\cdot5=30$.

Solution 2:

I have an approach that does not use graph theory and is pretty elementary.

You have numbers 1 to 6 which have to be assigned to the faces of the cube. Imagine you are facing the cube. You pick one of the six numbers and put it on the face nearest to you. This first number you assign does not matter in your count because of the symmetry of the starting configuration. However, the next number you assign has to be accounted for because the remaining five sides are not equivalent.

Consider the face opposite the face where you put the first number. This face can be filled in 5 ways.

The configuration after putting the second number is again symmetric because the four remaining faces are equivalent. Therefore, you can put any one of the remaining four numbers onto one of the remaining four faces as they are all equivalent. However, consider the face opposite the third number. This face can be filled in 3 ways.

Now you have only 2 more numbers, which can be filled in 2 ways to give two distinct arrangements.

The total number of arrangements = 2*3*5 = 30.

Solution 3:

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 $$[x_1 x_2\cdots x_6]Z(G)(x_1, x_2, \ldots x_6) = \frac{1}{24} [x_1 x_2\cdots x_6] (x_1+x_2+\cdots+x_6)^6 = \frac{1}{24}\times 6! = 30.$$