1. ## PLLs

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.

2. Originally Posted by JohnnyA
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.

3. Originally Posted by Lucas Garron
Originally Posted by JohnnyA
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.

4. 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.

5. Originally Posted by 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.
Indeed, I got that the wrong way around in my post.
But the point was not to reveal all the case symmetries.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•