# PLLs



## JohnnyA (Mar 1, 2010)

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.


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## Lucas Garron (Mar 1, 2010)

JohnnyA said:


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


72 is right.

You've just counted the H-perm 4 times, once for each AUF symmetry. Count that up for each PLL (including solved), and you get 72 from 22.


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## Tim Reynolds (Mar 1, 2010)

Lucas Garron said:


> JohnnyA said:
> 
> 
> > That gives 72 PLLs but that's far too many.
> ...



Wait, doesn't H only get counted once? It's things like T that are counted four times.


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## qqwref (Mar 1, 2010)

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.


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## Lucas Garron (Mar 2, 2010)

qqwref said:


> 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.


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