# Permutations and Combinations.



## jackolanternsoup (Jun 16, 2009)

I recently touched on this in my school syllabus and instantly decided to start wondering how the people who calculated the answer 43 quintillion permutations (of a Rubik's cube), managed it. 

So with all my knowledge, I got this far:

there are 8 corners. So total permutations 
= 8!
each corner has 3 orientations, but the last one depends on the other 7 so
= 3^7
there are 12 edges, hence 
= 12!
Again orientations of 12 edges-1 therefore 
= 2^11

Blah blah and there's 
8!*3^7*12!*2^11 which equals 8.65*10^19 (double the supposed number)

and after wiki-ing i found that they, for some reason, got everything the same but they decided to divide the number of possible edge permutations by two. 

Why is that?

[edit]
There are 12!/2 (239,500,800) ways to arrange the edges, since an odd permutation of the corners implies an odd permutation of the edges as well.

that's the explanation that wiki gave but that makes no sense to me sadly. Heelp


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## Stefan (Jun 16, 2009)

Permutation parity.


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## jackolanternsoup (Jun 16, 2009)

Woah that was fast thanks 

[edit]

now I'm wondering how to count probability of cases. It's easy-ish to count probability of a skip since there's only one possible position for those but how about for things like probability of Sune (out of the other 7 possible corner combinations or orientations), or the Z perm?

(could this get BUMPed thanks  )


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