# 3x3 Permutations?



## ChickenWrap (Jan 28, 2014)

I know other threads have been made on this topic, but none of them quite answered my question. Today, I wanted to try and determine the possible permutations for a 3x3, and got this far:

-	3 orientations for each corner, so 3^7 (since the last corner’s position is determined by the preceding 7)
-	8 edges can be permuted in one of 7 remaining places, so 8^7
-	12 edges can be flipped in 2 different orientations, giving 12^2
-	12 edges can be solved in one of 12 spots, giving 2^11 (since the 12th edge will be solved)

So my calculation was 3^7 * 8^7 * 12^2 * 2^11. Obviously, this is wrong....did I make a small error, or am I an idiot? Please explain where I went wrong, if possible!


----------



## TheNextFeliks (Jan 28, 2014)

Let's see:
Corners: 3 orientations 3^7 [last depends on others]
8 permutation positions 8!/2 (no parity) 
Edges: 2 orientations 2^11 (again final is dependent)
12 permutation positions 12!/2 (no parity)

This is definitely on the forums and even on Wikipedia.


----------



## ChickenWrap (Jan 28, 2014)

Thanks, I understand this except....why do you divide by 2 for the permutations?


----------



## kcl (Jan 28, 2014)

ChickenWrap said:


> Thanks, I understand this except....why do you divide by 2 for the permutations?



You can't swap an even number of pieces.


----------



## SpicyOranges (Jan 28, 2014)

kclejeune said:


> You can't swap an even number of pieces.


T perm? Y perm? J perm? L perm? V perm? Ra perm? Rb perm? all swap 4 pieces. 4 is an even number.


----------



## kcl (Jan 28, 2014)

SpicyOranges said:


> T perm? Y perm? J perm? L perm? V perm? Ra perm? Rb perm? all swap 4 pieces. 4 is an even number.



You smartass 

Ok, you can't swap an odd number of pairs.


----------



## IcyBlade (Feb 3, 2014)

U-perm, A-perm.

You divide by 2 because there can't be parity. There is only 1 place for the last two edges.


----------



## Kirjava (Feb 3, 2014)

IcyBlade said:


> U-perm, A-perm.



They are both an even number of 2-swaps.


----------



## Stefan (Feb 3, 2014)

kclejeune said:


> You smartass
> 
> Ok, you can't swap an odd number of pairs.



U A-perm.

Swaps an odd number of pairs (and does a four-cycle).


----------



## Renslay (Feb 3, 2014)

http://www.speedsolving.com/wiki/index.php/Laws_of_the_cube
"Only an even number of cubie swaps is possible"

With different words, the signature of the edge-permutation must equal to the signature od the corner-permutation.
See http://en.wikipedia.org/wiki/Parity_of_a_permutation

And yes, U-perm and A-perm are both possible only with *even* number of swaps, for example:
The permutation UL -> UB -> UR -> UL (an U perm) equals to UL-UB swap plus UL-UR swap (which is *two* swaps).


----------



## rj (Feb 3, 2014)

kclejeune said:


> You smartass
> 
> Ok, you can't swap an odd number of pairs.



Sometimes I wonder how the [censored word] you're so fast. You don't even know PLL!


----------



## kcl (Feb 3, 2014)

rj said:


> Sometimes I wonder how the [censored word] you're so fast. You don't even know PLL!



lol I know PLL. I was ridiculously tired when I wrote that and what I meant to say came out wrong XD


----------



## CompanionCube (Mar 1, 2014)

ChickenWrap said:


> I know other threads have been made on this topic, but none of them quite answered my question. Today, I wanted to try and determine the possible permutations for a 3x3, and got this far:
> 
> -	3 orientations for each corner, so 3^7 (since the last corner’s position is determined by the preceding 7)
> -	8 edges can be permuted in one of 7 remaining places, so 8^7
> ...



you are right but your calculations are wrong


3 orientations for each corner, so 3^7 (since the last corner’s position is determined by the preceding 7) 

8 edges can be permuted in one of 7 remaining places, so 8^7

8 edges permutation is 8!(this is just a different way of showing 8*7*6*5*4*3*2*1) divided by 2 (you divide it in two to avoid parity)

12 edges can be flipped in 2 different orientations, giving 12^2

the orientation for 12 edges should be 2^11 

12 edges can be solved in one of 12 spots, giving 2^11 (since the 12th edge will be solved)

the permutation of edges should be 12! divided by 2


----------

