Are the Permutations of Rubik's Cubes Combinatorics?

[DavidSanders]

I have done some research on combinatorics and Rubik's cubes for my extended essay for school and I cannot figure out if they are related. Some people say combinatorics include Rubik's cubes and other like puzzles, but the descripiton of combinatorics on Wikipedia does not seem to fit Rubik's cubes. Could someone please explain to me how combinatorics includes Rubik's Cubes, or what area of math Rubik's cubes are? And by the math I mean the equations for the possible number of permutations.

 

[spdqbr]

Typically people run to group theory when they think "Mathematical branches that are easily applicable to Rubik's Cube." Combinatorics is useful in puzzle theory, at the very least, for calculating number of possible states of a puzzle...

 

[frogmanson]

People use burnside's lemma to calculate the number of perms on cube.

 

[pwndnoobcuber]

i think the general formula for positions on any cube is:
E!/2 * 2^(E-1) * C! * 3^(C-1) * M!
where C is no. of corners
E is no. of edges
and M is the number of center pieces
i think it is wrong because i don't think it encorporates parity or opposite outer edges being swapped on the same 'multi-edge' like on a 7x7
i know its correct for a 3x3

 

[cmhardw]

i think the general formula for positions on any cube is:
E!/2 * 2^(E-1) * C! * 3^(C-1) * M!
where C is no. of corners
E is no. of edges
and M is the number of center pieces
i think it is wrong because i don't think it encorporates parity or opposite outer edges being swapped on the same 'multi-edge' like on a 7x7
i know its correct for a 3x3

You have a good start, but I agree that your formula does not work (except for 3x3). What stands out to me the most can be illustrated by the following example: try to use your formula for the 2x2x2.

For a 2x2x2 there are:
C=8 corners
E=0 edges.
M=0 centers

Using your formula I would get:
0!/2 * 2^(0-1) * 8! * 3^(8-1) * 0! = 22044960

Which is exactly six times as large* as the actual number. There will be a somewhat related problem with trying to use your formula on a 4x4x4.

Do you see the pattern of what I am trying to point out here? Please don't take this as a flame post, but rather as constructive criticism as to how you are on the right track. Your formula only needs more thought, and some tweaking, to be made correct.

*
--edit--
Bolded part is an edit for correctness. I did not give the right factor before.
--/edit--

Chris