Unlocking the Rubik Cube 2x2 Mystery

[Feynmanfan]

Hello everybody!

What's the number of possible configuration of a Rubik cube 2x2 (allowed movements are like in a 3x3 cube but here there are not central pieces that don't move).

THanks for your help

 

[Sirus]

Smells like homework, which means we need to see your work.

 

[Feynmanfan]

well, sorry for that.

The first thing I thought was 8!3^8 since there are 8 vertices and 3 possible positions for each one. But obviously in this answer there are repeated configurations.

If I prove that any permutation can be made in a 2x2 cube, then 24! would be the number of permutations but colours are repeated so it would be
24!/(4!)^6.

I'm a bit lost in combinatorics.

Thanks for helping me out.

 

[NateTG]

The first thing I thought was 8!3^8 since there are 8 vertices and 3 possible positions for each one. But obviously in this answer there are repeated configurations.

I don't understand what you mean by obviously there are repeated configurations. Can you give an example of an arragement of faces that this example repeats?

 

[FulhamFan3]

If I prove that any permutation can be made in a 2x2 cube, then 24! would be the number of permutations but colours are repeated so it would be

At first i thought you could do that. However that's assuming all the colors have complete freedom on the cube. On a cube every corner has 3 colors that can't be shuffled. I don't know how to account for that but I'm pretty sure the standard calculation for the number of permutations isn't valid.

 

[NateTG]

Really a bunch of this depends on what kind of stuff you're looking into.

For example, let's say that instead of moving the cube around, you're swapping stickers on the face of the cube.
Then, if each of panels were distinct you could have 24! possible combinations. However, there are 4 of each color on the cube, so, really it's
$$\frac{24!}{\left{4!}^6}$$
combinations. Since you can freely rearange the faces of each color.

Now, let's take a look at an actual cube:
Naively, the cube has eight moving parts - one for each vertex, and each of them can be in one of eight positions. Each of these positions has three possible orientations, and all of the vertices are distinct, so there are at most
$$8! \times 3^8$$
positions.
The actual number of positions that a mini-cube can achieve is somewhat smaller than that.

 

[learningphysics]

Really a bunch of this depends on what kind of stuff you're looking into.

For example, let's say that instead of moving the cube around, you're swapping stickers on the face of the cube.
Then, if each of panels were distinct you could have 24! possible combinations. However, there are 4 of each color on the cube, so, really it's
$$\frac{24!}{\left{4!}^6}$$
combinations. Since you can freely rearange the faces of each color.

Now, let's take a look at an actual cube:
Naively, the cube has eight moving parts - one for each vertex, and each of them can be in one of eight positions. Each of these positions has three possible orientations, and all of the vertices are distinct, so there are at most
$$8! \times 3^8$$
positions.
The actual number of positions that a mini-cube can achieve is somewhat smaller than that.

I think the correct answer is $$7! \times 3^8$$. Divide by 8, so that we don't include a reorientation of the cube.

 

[NateTG]

I think the correct answer is $$7! \times 3^8$$. Divide by 8, so that we don't include a reorientation of the cube.

If you fix a corner, it's pretty easy to see that the number is capped at $$7! \times 3^7$$. It would not be surprising to find out that the actual number is $$7! \times 3^6$$. This is well within the storage capacities of todays computers, so brute force is a viable approach to confirming this notion.

 

[Problem+Solve=Reason]

Can someone please explain 8!3^3. Thank you for helping out a rook...

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