# Calculating permutations for a 5x5 (almost)supercube



## sgtjosh (Apr 16, 2014)

Hey guys, 

I have an interesting problem I was hoping the greater minds here could help me with.

I'm wondering how I'd go about calculating the number of possible permutations for a 5x5 picture cube... Except this 'super cube' only has pictures on five of its six sides (The sixth side is just a solid color).

So the center pieces of five sides have to go into the same place every time, but the center pieces of the sixth side do not.

Any help would be greatly appreciated... And the more detail, the better!

Thanks a lot,
-Josh


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## Jakube (Apr 16, 2014)

We get: 

8! - possible corner permutation
3^7 - orientation of corners (notice 7 instead of 8, because the last corner orientation is forced by the others)
12! / 2 - midge-permutations, (divide by 2, because the last two midges are forced. It isn't possible too switch two midges alone)
2^11 - midge-orientations (last orientation is forced by the other 11)

=> 8! * 3^7 * 12! * 2^11 / 2

24! - 24 wings, every permutation is possible (it is possible to switch 2 wings without any other pieces, thanks too the face with only 1 color)

We have 2 groups of center pieces: 
Group 1: x-centers
Every permutation can be described with only the picture pices, the other 4 will be forced. 
=> 24! / 4!
Same with the t-centers: 24! / 4!

=> There are 8! * 3^7 * 12! * 2^11 / 2 * 24! * 24! / 4! * 24! / 4! = 17934974117270551244186247903990858235542284305932202306560963920617537536000000000000000 = 1.79 × 10^88


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## sgtjosh (Apr 16, 2014)

Dude, you're awesome. This is exactly what I was hoping to find out.

Thanks so much!


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## Jakube (Apr 16, 2014)

sgtjosh said:


> Dude, you're awesome. This is exactly what I was hoping to find out.
> 
> Thanks so much!



Oops! I made a little mistake. I edited the post above. There are 24! / 4! permutations of each center group, not 20!.


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## sgtjosh (Apr 16, 2014)

Jakube said:


> Oops! I made a little mistake. I edited the post above. There are 24! / 4! permutations of each center group, not 20!.



Awesome. Thanks a lot!


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## sgtjosh (Apr 21, 2014)

Hey there Jakob -

Just out of curiosity, how many permutations would be possible on a _7x7_ cube with the same type of sticker arrangement?


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## cmhardw (Apr 21, 2014)

Jakube said:


> We get:
> 
> 8! - possible corner permutation
> 3^7 - orientation of corners (notice 7 instead of 8, because the last corner orientation is forced by the others)
> ...



The centralmost centers must be rotated too. I don't see this in your calculation.



sgtjosh said:


> Hey there Jakob -
> 
> Just out of curiosity, how many permutations would be possible on a _7x7_ cube with the same type of sticker arrangement?



You have the explanation from the 5x5x5 cube of the same type.

For the 7x7x7 supercube with no pictures (and no logos) on the 6th side:

Corners and centralmost edges:
\( 8!*3^7*\frac{12!}{2}*2^{11} \)

Wings:
\( (24!)^2 \)
This calculation for wings uses the same reasoning as for the 5x5x5, except with 2 wing orbits now.

Centers:
\( \left(\frac{24!}{4!}\right)^6*4^5 \)

This centers calculation uses the same method as Jakube used only now there are 6 center piece orbits. The 4^5 accounts for the centralmost center pieces since they can noticeably twist in place on 5 sides of the 6.

Putting it all together:
\( 8!*3^7*\frac{12!}{2}*2^{11}*(24!)^2*\left(\frac{24!}{4!}\right)^6*4^5 \)

Using wolframalpha this is approximately:

*5.0896051928992777130079600798397771352708855555 × 10^204*


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## sgtjosh (Apr 21, 2014)

cmhardw said:


> The centralmost centers must be rotated too. I don't see this in your calculation.



Could that be fixed by just multiplying by 4 for every centerpiece whose orientation matters in the difference between a solved and unsolved state? (i.e. - 4^5 ?)

I really appreciate your help, guys!


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## cmhardw (Apr 21, 2014)

sgtjosh said:


> Could that be fixed by just multiplying by 4 for every centerpiece whose orientation matters in the difference between a solved and unsolved state? (i.e. - 4^5 ?)
> 
> I really appreciate your help, guys!



Yes, multiplying Jakube's result by 4^5 will give the number of combinations to the 5x5x5 almost supercube you described.


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## Jakube (Apr 21, 2014)

cmhardw said:


> The centralmost centers must be rotated too. I don't see this in your calculation.



Oops, completely forgot about it.


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