Message #1634

From: David Smith <>
Subject: Re: [MC4D] Further correction of erroneous order-2 Klein’s Quartic permutation count
Date: Sun, 01 May 2011 08:42:23 -0700

Hi everyone,

Argh!  Iit turns out there was a second mistake. :(  This one is especially embarrassing.  I apologize for these corrections, and in the future will be much more careful before posting my results.  It’s been way too long since I calculated permutations.  This stuff is about 100,000 times simpler than my n^d Rubik’s cube formula, which makes these corrections especially frustrating and embarrassing.  Hopefully I do not cast doubt on the validity of my Rubik’s cube formulas, which I checked for months, but I would understand if anyone questions them at this point.

I simply miscounted the number of different colors in the puzzle! I was using a count of 30 colors, when in fact the correct number is 24.  The really funny thing is that MagicTile *tells you* how many colors are in each puzzle, right next to the puzzle name!  So you can see my embarrassment. :(  Here is the (possibly?) correct number of permutations:

167!/(((7!)^23)*(6!)) =


If there are any more errors in this very simple calculation, I think I’m going to quit working on these!  Just kidding.  Not to sound arrogant (in fact I intend to convey the opposite) but this stuff is child’s play compared to my Rubik’s cube formulas. (Sigh.)

All the best,

— On Sat, 4/30/11, djs314djs314 <> wrote:

From: djs314djs314 <>
Subject: [MC4D] Correction of erroneous order-2 Klein’s Quartic permutation count
Date: Saturday, April 30, 2011, 9:32 PM


Hello all,

Aha! I did make a mistake, as I suspected I might have. It’s been too long since I counted permutations. :) My previous answer was off by a factor of 2; the 2 in the denominator should not have been there. I made the classic mistake of taking even permutations into account when there are identically colored pieces (in this case, stickers). Naturally, with identically colored pieces parity does not matter, as even and odd permutations are identical.

Here is the correct formula and number of permutations:

209!/(((7!)^29)*(6!)) =


My apologies for not catching this sooner; I’ve been very busy today with emails. I am virtually certain that this count is correct, as I just sent Ray a complete derivation of it (which is how I discovered my error).

Thanks to everyone for your patience, and I’ll keep you informed of my progress.

All the best,