# Message #1635

From: David Smith <djs314djs314@yahoo.com>

Subject: Re: [MC4D] Further correction of erroneous order-2 Klein’s Quartic permutation count

Date: Sun, 01 May 2011 09:50:03 -0700

Hi everyone,

There was a third, major mistake. It seems like I need to take a break, at least from posting erroneous results. This is very frustrating, and certainly not as easy as I thought it was. I didn’t realize this more subtle error (than the other two) until I was taking a look at the order-3 puzzle. Half an hour is hardly sufficient, especially when I am new to calculating permutations of this type of puzzle, to have any confidence in my answers!

My sincerest apologies, and perhaps I’ll collaborate with Roice, if he is able, willing, and has the time, to make sure my counts are accurate before posting any more results.

I’m sorry, and will stop flooding the message board with results, until I am certain beyond any doubt they are correct.

All the best,

David

— On Sun, 5/1/11, David Smith <djs314djs314@yahoo.com> wrote:

From: David Smith <djs314djs314@yahoo.com>

Subject: Re: [MC4D] Further correction of erroneous order-2 Klein’s Quartic permutation count

To: 4D_Cubing@yahoogroups.com

Date: Sunday, May 1, 2011, 11:42 AM

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!)) =

1425599937004511574183572646332129723473495272508015637852647635847

6905825301087025100107412409270064909992900731302354053195964246823

6192752417277237699906089211520063444550594267719476697558665102816

164056186968

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,

David

— On Sat, 4/30/11, djs314djs314 <djs314djs314@yahoo.com> wrote:

From: djs314djs314 <djs314djs314@yahoo.com>

Subject: [MC4D]

Correction of erroneous order-2 Klein’s Quartic permutation count

To: 4D_Cubing@yahoogroups.com

Date: Saturday, April 30, 2011, 9:32 PM

```
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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!)) =

2982258374829406765091841007287208110044231115104792401778764352522005836340463837850121874947185190056715803101607827209655156207636181402167465898185982442986345576457227559003601279297679332452494926096129531568368614235884567287678429277548260303941234016334643200000000000000000000

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,

David