Lower bounds on numbers of arrangements of a Rubik's cube Last night, a friend of mine informed me that there were forty-three quintillion positions that a Rubik's Cube could be in and asked me how many there were for my Professor's Cube (5x5x5).
So I gave him an upper bound:
$$8!~3^8\cdot24!~2^{24}/2^{12}\cdot12!~2^{12}\cdot24!/4!^6\cdot24!/4!^6,$$
did some rough approximation, and said it was around 10^79.
Then I decided I'd try to give a lower bound, and came up dry.
In the Wikipedia article, we read that:


*

*The orientation of the eighth vertex depends on the other seven.

*The 24 edge pieces that are next to vertices can't be flipped.

*The twelfth central edge's orientation depends on the other eleven.

*The parity of a permutation of the vertices and of the edge pieces are linked.


The second point is easy to see by imagining the route that such an edge piece takes under all the various moves. (The Wikipedia article opts to give a mechanical reason instead, which is a questionable practice.) The other points are not too hard to see either.
So we divide by $3\cdot2^{24}\cdot2\cdot2$ and get the answer, around $2.83\cdot10^{74}.$
My friend does not know group theory, and the only proof I know of the independence of the rest of the stuff uses group theory to a considerably greater extent than that which proves the dependent parts dependent. Can anyone think of a simpler proof of a reasonable lower bound? One that, say, at least puts the number greater than that for Rubik's Revenge (4x4x4), which is about $7.40\cdot10^{45}$?

Chris Hardwick has come up with a closed-form solution for the number of permutations for an $n\times n\times n$ cube: http://speedcubing.com/chris/cubecombos.html
