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PLLs

JohnnyA

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Dec 19, 2008
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How do I go about the calculation of the number of PLLs? This is for a school essay. I've done this:

4!x4! to give the total number of possible permutations.
4!x4!/2 to get rid of the impossible ones.
4!x4!/2/4 to get rid of symmetry.

That gives 72 PLLs but that's far too many.
 

Lucas Garron

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H and solved get counted once; each N perm gets counted once each; Z and E get counted twice. That leaves 64 positions. Each of the remaining PLLs (that's 16 of them) are counted 4 times each.
Indeed, I got that the wrong way around in my post.
But the point was not to reveal all the case symmetries. :p
 
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