Why do we divide by 12 when calculating the number of different instances of the Rubik's cube?

Question

1. $\frac{8!\cdot3^8\cdot12!\cdot2^{12}}{3\cdot2\cdot2}$ This is the solution, do you know a paper in which this formula is explained, why we calculate like this?

2. Then I heard that the numerator of the fraction would be the result if one did not consider the mechanics of the cube, but could simply assume that all 54 faces of the cube could take on any of the available colours. But then what about the centre pieces? If we ignored the mechanics of the cube, could they also take on any colour, if yes, how does the numerator take the centre stones into account?

Answer 1

Just rotating the cube as one complete object gives you 24 ways to put it in front of you (say one a designated square on the table). These rotations are not seen as "moves", so the formula does not account for them.

This means that in any state of the cube, you can always put it in front of you with the center pieces in a standard orientation. And when manipulating the cube you can, if you wish, always keep the center pieces fixed and rotate other planes around it (although that may not be how cubists do the solving).

If we would also count the 24 "placements" of the cube as different states then the formula would give an even bigger number, 24 times bigger. But since that is not the convention we assume the center pieces fixed, and then note that we potentially have $8!$ permutations of the corners, $3^8$ orientations of the corners, $12!$ permutations of the edges, $2^{12}$ orientations of the edges. And then divide by $12$ because only $1/12$th of all those potential possibilities can be reached.

Only 1/3rd of the corner rotations can be reached without physically taking out one corner and putting it back in a different orientation, only half of the orientations of the edges without taking one out and flipping it, and only even total permutations of their positions are possible. You can see that by looking at each elemental move (90 degrees rotation of one of the faces, center pieces fixed).

You'll see that such a move always gives an odd permutation of 4 edges and an odd permutation of 4 corners (for both a 4-cycle) so the total sign remains even. Also you can define a parity for the edge orientations and show that it never changes (leaving half of the possibilities). And a kind of "modulo 3" parity for the twists of the corners, which will always change by 3, leaving just one third of those possibilities. The cubology literature can provide you with more detail how to define this all, but this is the story.

Answer 2

Ok so based on your questions, let's try to clear something up first:

When people talk about the number of permutations of a Rubik's Cube, they do NOT include rotating the whole cube. So, for example, let's say you had a solved cube. Hold it with the white side on top. Now rotate the whole cube so that the blue side is one top. These are not considered different permutations. Only turning a face of the cube will result in new permutations. Based on this fact, we can say that the center squares do not move, because face turns do not move center squares.

So to more directly answer your questions, the numerator of that expression (ie without dividing by 12) would be the number of permutations if we were allowed to disassemble the cube into its individual pieces, then reassemble it in any way we want. When you take apart a cube, notice that there are corner pieces and edge pieces (remember, the center pieces do not move, so we can ignore them). These pieces cannot be broken down further. That is why the numerator does not represent the number of permutations on all 54 squares. It only represents the number of permutations on the individual pieces.

So where does the 12 in the denominator come from? Well, let's go back to disassembling the cube. If we take apart the cube, scramble the pieces, and put it back together in a cube shape, we'll get a very scrambled cube. If we try to solve this cube, there's a good chance it's impossible! This is because if we start with a solved cube, and do normal turns to it, there are some positions we just cannot reach.

There are three basic types of positions that are unreachable:

1. One edge is rotated.
2. One corner is rotated.
3. Two edges are swapped.

For 1., say you take a solved cube, then pop out one edge piece, turn it 180 degrees, then pop it back in. This position is impossible to reach. To put it another way, say we disassemble the cube, scramble the pieces, and start putting it back into a cube shape. We can put in the first 11 edges in any rotation we want. But when we're putting that last edge piece in, there is only one right way to do it so that we're left with a solvable cube. So this is where a factor of 2 comes from.

For 2., it's the same argument as for 1., now for the corners. There are 3 ways to put in a corner, and only one of them leads to a solvable cube. So this is where a factor of 3 comes from.

For 3., now we are talking about the position of the edges. For 1. and 2., the constraint was on the rotation of the last edge or corner. But now we're talking about the position of the last two edges, ie where they are placed (not caring about how they are rotated). So same as before, as we're putting our scrambled pieces back into a cube shape, we can put the first 10 edges in wherever we want, until we get to our last two edges. Here there is only one right way to do it so that we're left with a solvable cube. This is where a factor of 2 comes from.

So putting all the factors together, we have $2 \cdot 3 \cdot 2 = 12$.