Mathematics question on total permutations

[Silky]

As almost all of you are aware the standard Rubik's Cube has ~ 43 quintillion permutations. My question is, is this calculation unique permutations ? If not what are the total unique permutations.

 

[Spacey10]

Yeah they are all different
There are more, but they are impossible states (corner twists, 2 edge swap, edge misoriented)

 

[2018AMSB02]

As almost all of you are aware the standard Rubik's Cube has ~ 43 quintillion permutations. My question is, is this calculation unique permutations ? If not what are the total unique permutations.

Yes, this is unique permutations, so all the different combinations of colors on specific sides. If you factor in center orientation and cube rotation, the number becomes 2,125,922,464,947,725,402,112,000

 

[s_e_a_n666]

As almost all of you are aware the standard Rubik's Cube has ~ 43 quintillion permutations. My question is, is this calculation unique permutations ? If not what are the total unique permutations.

Yes it is unique permutations. For 2x2 and 4x4 you have the problem of no fixed centers so you can have non unique permutations. To get to the number of unique permutations you divide by 24 because there are 24 orientations the cube can be held.

 

[xyzzy]

As almost all of you are aware the standard Rubik's Cube has ~ 43 quintillion permutations. My question is, is this calculation unique permutations ? If not what are the total unique permutations.

Depends on what you mean by "unique". (This is a recurring theme: almost every problem about counting things (i.e. combinatorics) becomes much easier once you have a clear idea of what exactly you're trying to count. Obviously, if you're trying to count something without even knowing what it is, that's not going to be an easy task.)

The 43 quintillion number comes from this: hold the core in a fixed orientation (e.g. white top green front), and count the number of ways the edge pieces and corner pieces can be placed around it so that it's solvable. This doesn't count any configuration multiple times, but only if you take "configuration" to mean the exact colours on each of the facelets.

For instance, this considers H perm on the U layer (set up with M2 U M2 U2 M2 U M2) and H perm on the D layer (set up with M2 D M2 D2 M2 D M2) to be distinct positions. Do you want to treat them as being equivalent instead? If you do, figure out exactly which kinds of equivalences you accept. Rotations? (All 24 rotations, or only some of them?) Reflections? Inverses?

Once you've decided on that, you can calculate the exact number of configurations with given symmetry/antisymmetry subgroups (or just look it up on kociemba.org), then use not-Burnside's lemma to calculate the number of equivalence classes (with respect to your chosen sense of "equivalence").