PLLs

[JohnnyA]

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]

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.

[Tim Reynolds]

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.

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

[qqwref]

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.

[Lucas Garron]

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.