Ignoring the solved case and counting mirrors together but inverses 

First off, I looked at the number of totally distinct cases, that'd be 
6!/2 = 360. Exploiting two planes of symmetry that gives 90 cases 
while exploiting 4 planes of symmetry gives 15 cases.

But that was just to get a rough estimate so that I know that my 
answer is going to be between 14 and 79. Some cases are self-mirrors 
that shouldn't be counted twice and some cases have more or less 
symmetry than others... a bit of a mess trying to apply Burnside's 
Lemma on.

So then I broke it up into categories:
There are 8 cases involving a 5-edge cycle.
There are 7 cases involving a 4-edge cycle and a 2-edge cycle.
There are 6 cases involving two disjoint 3-edge cycles.
There are 4 cases involving a single 3-edge cycle.
There are 6 cases involving two disjoint 2-edge cycles.
There is the solved case.

Thus I conclude that there are 31 cases not counting the solved case. 
Also 3 of the algs needed are from PLL, so it's really only 28 algs to 
memorize, some of which maybe inverse of each other.

That is a quick count, nothing rigorous... see if you can find any 
flaws in my reasoning.


--- In zbmethod@yahoogroups.com, "Mike Bennett" <[EMAIL PROTECTED]> 
> Can someone help me work out the number of possible combinations for
> when there are 1x3x4 blocks on the sides, all four U layer corners
> correct, and the final 6 edges and centers oriented?  I'd really like
> to know, to see if this is a strategy worth pursuing.

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