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Could anyone explain why the number of legal or reachable combinations of a $3\times 3\times 3$ Rubik's Cube is $1/12\mbox{th}$ of the total. I understood the logic behind the total number of combinations: $8! \cdot 2^{12} + 3^{12} \cdot 12!$. What I am not able to understand is why do we divide this quantity by $2 \cdot 3\cdot 2$. I don't have any idea about parity or group/set theory. Is there another possible explanation for it?

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The short answer is that there are certain relationships which must be met:

The orientation (ie how it is rotated) of the last corner is determined by the orientations of the first 7 corners. Stated another way, given the orientations of the first 7 corners, the orientation of the last corner is set. So instead of 3 possible orientations for the last corner, there is only 1. So we must divide by 3.

The orientation of the last edge is determined by the orientations of the first 11 edges. So for the last edge, instead of 2 possible orientations, there is now only 1. So we must divide by 2.

The last restriction has to do with the permutation of the cube (ie, where the pieces are located, ignoring orientation). So, as it turns out, it is impossible permute just 2 edges and it is impossible to permute just 2 corners, but it is possible to permute 2 edges and 2 corners together. Put another way, the permutation of the corners determines the parity of the permutation of the edges. So we must divide by 2.

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This is because even though it is possible to pick a cube apart and put it back again however you want it to be, the set of legal moves (turning a side) is somewhat limited. There is for instance no combination of moves with net result of turning just one corner cube one third of a revolution. the $1/12$ is there because it is possible to find a maximum of $12$ "independent" states, none of them reachable from any of the others (one of them being reachable from a solved cube). Other than that, a cube of any of these arrangements behaves like a normal cube, so by symmetry, a twelfth of all combinations is reachable from each of them.

Numberphile has a video on the cube, they spend a minute or two explaining where the $1/12$th comes from (without going into detail about why exactly it is $12$).

There is an analogue to the classical sliding puzzle. If you look up the 15 puzzle, you can see that the number of reachable states in a sliding puzzle is half the total number of permutations of the tiles. Again, this is caused by a limitation on the legal moves (sliding up or down).

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