Method question

[Erik]

Ok, I've just been playing around with some other ways of solving the cube. But now I need to know how many cases there are for a particular step, but I'm horrible at such mathematical stuff.
Suppose you had to solve the edges of the top layer and the right layer like this:

How many cases would it be to solve them all at once? And with orient then permutate?

 

[Joël]

For all seven edges: 7!*2^6 / 2 = Too many cases to do in one step (about a million. But if you rotate the cube, you might have less cases, because you can solve some 'different' cases with the same alg after y2 z .

Orient first would be easier, I suppose, allthough there are still 6! = 720 cases for the permutation step...

 

[mmwfung]

Isn't it 6!*2^7 / 2 cases instead of 7!*2^6 / 2? About a million??? 6!*2^7 / 2 = 46080 and 7!*2^6 / 2 = 161.280.

Michael Fung

 

[Erik]

Maybe that's a bit too much