# How many unique permutations exist on this type of rubik's cube?

NOW WAIT,I am not asking about a normal rubik's cube, I know there are 43 quintillion unique scrambles for the regular cube, but what I am asking goes beyond that.

I am asking, how many unique possible colour patterns can exist on a stickered rubik's cube if we consider the possibilities of:

1. Twisting corners and flipping edges

2. Peeling off stickers and applying them back in different colour schemes for eg. blue and green can be next to each other, white and orange opposite etc.

I want the answer to include even the combinations of the two things I mentioned above, for eg. in a colour scheme of white being opposite to orange and THEN twisting corners and flipping edges. The same being applied to all possible colour schemes.

(also, if you can recommend a better title for the question that isn't as ambiguous as it is now and implies that I am not referring to a regular rubik's cube without being too long, pls do so)

Also, I realize that this question is very similar to this previously asked one:

How many disconnected graphs of the Rubik's cube exist?

But I don't think it includes flipping edges and twisting corners, which is why I asked this question

EDIT: By 'Peeling off stickers', I mean you can physically peel off any of the existing 54 stickers divided into 6 groups of 9(by colour) and stick it back onto the cube on any of the sides of the cube as a whole unit(of colours, so I can for eg. transfer all 9 white stickers and paste them on the green side while putting the green stickers on the white side) and then include the resulting possible scrambles of that colour scheme. To clarify, a "colour scheme" means the orientation of colours in relation to each other. For example, the standard colour scheme in a rubiks cube is green opposite to blue, yellow opp. to white, red opp. to orange and having green on the front facing side of the cube and orange on the right side of the cube when the yellow side is facing upwards. In my question, you can have the colours oriented however you like as long as its possible,like having yellow ADJACENT to white which is usualy not the case.

• Do you know about finite group actions and Burnside’s Lemma? May 21 at 19:01
• No, I have no idea what that is May 21 at 19:02
• Number 2 "Peeling off stickers..." is not clear... can I peel off the stickers and put them back in any way I want? Or only certain ways? "Color schemes" sounds like there is some kind of restriction.
– Ted
May 21 at 19:41
• I'm guessing that the "peeling off stickers" rule only allows us to completely interchange colors, right? In other words, I could make all the red stickers white, all the white stickers green, and all the green stickers red, but I couldn't make it so that all the centers are blue and one of the corners is completely orange. Am I understanding that right? May 21 at 19:55
• I have edited the question to address your queries, let me know if anything else requires clarification. May 21 at 20:23

So I'm not totally sure what you're asking. If it is simply "How many configurations exist if I can peel off the stickers and put them anywhere?" then the answer is the multinomial coefficient

$${54 \choose 9,9,9,9,9,9} = 101097362223624462291180422369532000000$$

which is approximately $$1.01\times10^{38}$$.

But that would make your #1 redundant, so I think what you're saying is that the cube must start with only one color per side, and then you can scramble it up however you want by taking the cubies apart and putting them back together.

To answer this question, first consider your #1. Relaxing this constraint means there are 12 times more configurations than a regular Rubik's cube.

Then consider your #2. We have to figure out how many unique ways there are to put six colors on to six sides. An example of something that is NOT unique is take a regular Rubik's cube, and then apply a whole-cube rotation (X, Y, Z in the usual notation). Are there different colors on different sides now? Yes... but that's not really a unique cube.

So WLOG let's put white on the Down side of the cube. Now consider the Upper side. Each of the five remaining colors would result in a different unique configuration, so put any remaining color there (say green). Therefore so far we have $$5$$.

Now consider the Left side. It doesn't matter what color we put here, as none of them will result in a different unique configuration, so put any remaining color there (say blue). So far we still have $$5$$.

Now consider the Right side. Each of the three remaining colors would result in a different unique configuration. So put any remaining color there (say yellow). So now we have $$5\times3$$.

Now consider the Front side. Each of the two remaining colors would result in a different unique configuration. So now we have $$5\times3\times2$$. And there's only one remaining color to place on the Back side, so it must go there.

So $$5\times3\times2=30$$. That is for your #2.

Putting it all together we have part 1 times part 2 times the number of regular Rubik's cube permutations:

$$12\times30\times43252003274489856000=15570721178816348160000$$

or approximately $$1.55\times10^{22}$$.

Another way to calculate part 2 is to say there are six colors and six spots for them, so $$6!$$, but we have to divide by the number of ways to rotate a whole cube. To calculate the number of ways to rotate a whole cube, there are six different colors that can be on the Upper side, and once that is fixed, there are four colors that can be on the Left. So $$6!/6/4=30$$.

• There's one part I don't get, why did you say that it dosent matter whatever colour we get on the left side? Since from what I understand it does matter what colour is put there,if we put yellow there for example, that will result in a edge piece with green and yellow on it, but putting blue will give an edge with green and blue on it. So that means that the which colour goes on the left side would make an impact since the edge pieces change completely depending on what it on that side, thus giving completely new permutations of the cube. May 21 at 20:59
• Maybe I didn't to the best job of explaining it, but here we're trying to calculate how many unique ways we can put six colors on six sides. If you want, you can use the alternate explanation at the bottom. But to better explain the first method, say we already put white on the Down, and green on the Upper. With these two colors in place, there are only so many possible configurations left, call it $n$. Now put yellow on the Left side. There are still $n$ possible configurations. You can verify this by rotating the whole cube keeping UD the same, but you can put yellow anywhere. May 21 at 21:49
• ohh I think I get it now.There is one more problem though, I think there will be double counting in some cases, since, its possible that the effect of stickers changing can be nullified by twisting or flipping in an opposite way, but in your calculations I think that will be included. However other than that, I think your answer is perfect and answers my question, Thank you very much! May 21 at 22:25