# Message #17

From: David Vanderschel <DvdS@Austin.RR.com>

Subject: Fwd: 1+1=2, and…

Date: Fri, 01 Aug 2003 19:58:12 -0000

Date: Sat Feb 1, 2003 7:46 pm

— In MC4D@yahoogroups.com, p’tit bonhomme <ptit-bonhomme@t…>

wrote:

Sorry, It took me a while to find time

to answer, here is the way I did my calculus:

First, I’m sure, everybody would anderstand, what

I will call, the 1-colored, the 2-colored, the

3-colored and the 4-colored elements.

Let’s start with the 3x3x3x3, we can count

16 4-colored,

32 3-colored,

24 2-colored,

the 8 1-colored elements are immobile and will

allow us to locate the position of every other

element.

```
There is two steps, the first one consists to<br> count the possible permutations of the place of<br> the elements, the second one to count the possible<br> positions of the elements on their places.<br>
Not all the permutations of the 24 2-colored and the<br> 32 3-colored are possible, only the permutations that<br> have the same parity on the 2-colored and the 3-colored,<br> to check that, it is enough to check it on the basic moves<br> of one face. So we can count<br>
(24!x32!)/2
All the even permutations of the 4-colored, and the odd<br> ones are impossible, It can be checked on the basic moves,<br> so we count<br>
16!/2
The second step, now that all the positions are determined,<br> the 2-colored can have two positions on one place, but the<br> last one, whose position is fully determined by the positions of<br> the 23 others, so <br>
2^23
Every 3-colored can have 3! positions on one place, but<br> the last one, which can only have 3 positions, (see the attached<br> file: 3-Twist.log), so<br>
(3!)^31 x 3
Finally, the 4-colored, can have 4!/2 positions on one<br> place, but the last one, which can have only 4 position,<br> (see the attached file 4-Twist.log), so<br>
(4!/2)^15 x 4
```

All this counts, are independant, so the positions of the

3x3x3x3 is the product of all this numbers:

(24!x32!)/2 x 16!/2 x 2^23 x (3!)^31 x 3 x (4!/2)^15 x 4

There is some subtilities for the 4x4x4x4:

As there isn’t centers, we need to fix an element to locate

all the others, let’s fix a 4-colored (it can be done with

a 3-colored),

we have:

16 4-colored,

64 3-colored,

96 2-colored,

64 1-colored

The even permutations of the 4 colored are possible, so

(15!/2)

And they can have 4!/2 positions but the last, only 4, so

((4!/2)^14)*4

(Note that we have fix one of them, so it differs from the

counts of the 3x3x3x3)

This time, all the permutations are even for the 3-colored:

64!/2

(Note that on the 4x4x4, the 2-colored accepts odd permitations)

the 3-colored have 3 positions, on a place, and the last is fully

determined by the 63 precedent (Note that it differs from the 3x3x3x3,

and that two 3-colored that hold the same colors can’t be on a

same place, in the same position), so

3^63

The 2-colored accept only even permutations, but as they

are undistinguable, 4 by 4, and that we count the visually

different positions, we have

(96!/2)/((4!)^24/2) ( =(96!)/((4!)^24) )

Their positions: 2^95, because the position of the last is

determined by the others

The same problem appears for the visually different positions

of the 1-colored, so

(64!/2)/((8!)^8/2)

Finally, we have:

(15!/2)*((4!/2)^14)*4*(64!/2)*(3^63)*(96!/2)/((4!)^24/2)*(2^95)*

(64!/2)/((8!)^8/2)

You can verify that if we fix one of the 3-colored at the beginnig, we

have a different

formula, but the same result.

As I wrote all this, I realised an error in the calculus I send you

for

the 4x4x4x4,(just a factor 2)

, and I re-did all the calculus for the 5x5x5x5, because I forgot

whole

families of elements, I’m quite

sure about this figures, now that I put everything on paper.

(Note that for the 5x5x5x5, there different families of 1-colored,

2-colored and 3-colored:

1-colored: 1 group of 48

1 group of 96

1 group of 64

2-colored: 1 group of 24

1 group of 96

1 group of 96

3-colored: 1 group of 32

1 group of 64

4-colored: 1 group of 16)

English is not my first language, I hope I was clear enough.

If you have any comments, remarks or critics, don’t hesitate to make

me

know, that’s why the group

stands for.

2x2x2x2

> (15!/2)*((4!/2)^14)*4;

```
3 357 894 533 384 932 272 635 904 000
```

3x3x3x3

>(24!x32!)/2 x 16!/2 x 2^23 x (3!)^31 x 3 x (4!/2)^15 x 4 =

1 756 772 880 709 135 843 168 526

079 081 025 059 614 484 630 149

557 651 477 156 021 733 236 798

970 168 550 600 274 887 650 082

354 207 129 600 000 000 000 000

4x4x4x4

> (15!/2)*((4!/2)^14)*4*(64!/2)*(3^63)*(96!/2)*(2^95)/((4!)^24/2)*

(64!/2)/((8!)^8/2);

```
130 465 639 524 605 309 368 634 620 044<br> 528 122 859 025 488 438 611 959 323 482 221 544 701 493 566<br> 589 669 139 598 204 956 926 940 147 059 366 252 849 247 482<br> 898 636 104 705 417 194 760 866 897 307 590 845 202 461 293 <br> 100 468 293 214 262 958 591 194 739 437 727 430 945 469 384<br> 490 361 714 647 847 550 801 897 750 293 894 453 665 815 572<br> 829 257 758 907 425 128 919 808 862 616 259 604 997 210 112<br> 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
```

> (48!)/((6!)^8)*(96!)/((12!)^8)*(64!)/((8!)^8)*((24!*32!)/2)*((3!)

^31)*(2^23)*(64!/2)*(3^63)*(16!)*((4!/2)^15)*4*(96!)/((4!)^24)*(2^95)*

(96!)/((4!)^24)*(2^95);

```
82 438 037 949 266 001 798 818 537 185 591 <br> 872 622 513<br> 110 723 064 887 446 896 829 783 759 216 987 747 133 338 824 870 722 <br> 761 820 399<br> 091 803 906 672 200 562 788 191 831 782 678 757 916 210 500 720 119 <br> 109 924 738<br> 176 584 565 957 060 359 083 845 305 523 104 279 597 706 831 282 623 <br> 377 308 298<br> 270 256 110 577 915 550 842 311 947 852 455 908 640 926 513 887 950 <br> 693 259 734<br> 488 795 516 741 718 855 632 012 409 017 950 565 283 705 637 693 567 <br> 551 399 451<br> 022 890 300 760 696 806 001 691 690 503 354 312 640 767 127 338 809 <br> 808 328 091<br> 810 728 167 611 236 202 648 298 979 969 629 944 753 096 301 122 250 <br> 183 937 655<br> 748 970 939 083 829 108 821 970 975 167 712 732 490 661 498 153 951 <br> 649 064 753<br> 809 644 951 943 686 550 000 978 275 868 933 342 691 504 813 788 347 <br> 064 370 621<br> 775 923 549 337 026 399 778 184 629 950 873 600 000 000 000 000 000 <br> 000 000 000<br> 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 <br> 000 000 000<br> --- End forwarded message ---
```