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Given a 3x3x3 cube (like a rubik's cube) where each of the 27 cubes has a distinct number, how many unique permutations are possible? Simple rotations of the entire cube should not be counted.

The solution might be related to Rubik's cube and counting

Edit: to be clear about the the question I'll suggest an application.

I want to create a substitution cipher for english where I put the letters a-z and space, which is 27 characters, on each cube in a rubik's-type cube. I have a text I would like to encrypt. I have a starting orientation for the cube and I take the first letter and find it's location in the cube. I then rotate the cube 90 degrees in one direction and write down the resulting letter in that position. I repeat for all the letters in the message. I now have an encrypted message. To decode the message, I would have the same process, just in reverse. The process could be more complicated involving plane rotations and cumulative permutations if desired.

My question is how many different permutations of this cube could there be, that are not simple rotations of each other.

Community
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Yehosef
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  • See this answer on Wikipedia. – Ben Grossmann Jun 30 '14 at 22:28
  • Note that the pieces of a Rubik's cube can be permuted in such a fashion that it cannot be solved by combinations of rotations of the various faces. As such, there are actually two possible questions here: How many unique solvable permutations of a Rubik's cube exist? & How many unique solvable & nonsolvable permutations of a Rubik's cube exist? where solvable means that it can be solved by combinations of rotations of the various faces. – Pockets Jun 30 '14 at 22:37
  • wow - thanks @Omnomnomnom. I'm still not sure from that entry what the number is, though it's obviously very big. It seems that they are not excluding simple rotations or dealing with the fact that in a real rubiks cube a corner piece can never be put in the middle. In my question, that's a possibility. – Yehosef Jun 30 '14 at 22:37
  • just to be clear - I'm not asking for the permutations of a real rubik's cube, but a 3x3x3 set of cubes. And I'm not interested in "solving" the cube in my question, just knowing the number of possible unique permutations. – Yehosef Jun 30 '14 at 22:41

2 Answers2

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Edit: the answer is only valid for an actual rubik cube. The question has been clarified, making this question irrelevant.

You only have 20 movable squares on a Rubik's cube (8 corners and 12 borders) as the center squares cannot move relatively to each other.

The borders also cannot be swapped with corners.

Also if you consider the 6 centers immovable, there are no duplicate combinations, which is very useful.

Corners have 3 possible rotations, and borders have 2. Technically centers have 4, but that doesn't change anything.

All things considered, you have (!8 * 8^3) * (!12 * 12^2) possibilities.

Now, there are plenty of impossible rubik's cube. So if you take it all into account, and calculate that only 1/12 permutations are solvable, you get 43,252,003,274,489,856,000.

Lou Garczynski
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  • note, I'm not interested in a real Rubik's cube, just using it as an example. I want the numbers permutations where the value assigned to a corner, could be moved to the value assigned to an edge or the center. Does that affect your answer? – Yehosef Nov 30 '16 at 12:09
  • I don't understand your goal, can the smaller cubes move freely compared to each other? – Lou Garczynski Dec 04 '16 at 09:12
  • Sorry for the confusion - the "movement" is like a rubik's cube and for one permutation a corner would not move. But the value represented by on corner could be in a different piece (edge, center, etc) in a different permutation. I added a sample use case in my question. In that case where a certain letter, let's say "a" is in a corner. In a different permutation, "a" might be on an edge. Does that help? – Yehosef Dec 04 '16 at 09:52
  • Ok, so basically you can write any letter on any face, and then scramble? – Lou Garczynski Dec 05 '16 at 13:40
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First, we figure out how many arrangements there are, including "simple rotations". There are $3^3 = 27$ blocks to be freely permuted. This means that there are $27!$ possible arrangements of the blocks (including simple rotations).

Now, note that any given arrangement can be rotated to yield 6 $24$ (see comment below) superfically different arrangements. So, our count of $27!$ redundantly counts each rotated arrangement $24$ times. It follows that our desired total is $$ \frac{27!}{24} \approx 4.54 \times 10^{26} $$

Ben Grossmann
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