# Message #2104

From: Andrew Gould <agould@uwm.edu>

Subject: RE: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)

Date: Sun, 06 May 2012 16:35:37 -0500

The choice between 31 and 32 comes down to how you define the locations of

pieces. If you define all their locations relative to one of the pieces

it’s 31, but if you define what moves and what doesn’t for each twist you

can make it 32. I note that for 32, it would be tricky to say that rotating

the entire puzzle doesn’t change the state.

–

Andy

From: 4D_Cubing@yahoogroups.com [mailto:4D_Cubing@yahoogroups.com] On Behalf

Of charliemckiz@rocketmail.com

Sent: Wednesday, May 02, 2012 12:46

To: 4D_Cubing@yahoogroups.com

Subject: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)

2x2x2x2x2

(31!/2)(60^31) =

5453565517530819705863526338911096321376472677744640000000000000000000000000

0000000000000

approx. = 5.4 x 10^88

Should the 31 be 32?

Mine is

32!/2*(5!/2)^(32) =

1047084579365917383525797057070930493704282754126970880000000000000000000000

00000000000000000

approx. = 1.0 x 10^92