# Message #425

From: spel_werdz_rite <spel_werdz_rite@yahoo.com>

Subject: Discretions on the permutations of the 3^4

Date: Mon, 17 Sep 2007 03:58:29 -0000

Out of boredom, I chose to see for myself the number of positions on

the 3x3x3x3 and I kept getting half the value!

Here’s what I did:

24!x32!x16! - This is the total number of positions accounting all

n-sided stickers

(2^24)x(6^32)x(12^16) - The total number of orientations accounting

for all n-sided stickers

Now I divided numbers appropriately based on these findings:

2-I could not switch just two 2-colored stickers.

2-I could not switch just two 3-colored stickers.

2-I could not switch just two 4-colored stickers.

2-I could not flip just one 2-colored sticker.

2-I could flip just one 3-colored sticker, but only in three

orientations, not six.

3-I could flip just one 4-colored sticker, but only in four ways, not

twelve.

So my end result was:

((24!/2)x(32!/2)x(16!/2))x((2^24)/2)x((6^32)/2)x((12^16)/3)=

878,386,440,354,567,921,584,263,

039,540,512,529,807,242,315,074,

778,825,738,578,010,866,618,399,

485,084,275,300,137,443,825,041,

177,103,564,800,000,000,000,000

Which is half of the currently accepted value. I think the argument

lies where Eric Balandraud stated "Not all the permutations of the 24

2-colored and the 32 3-colored are possible. Only the permutations

that have the same parity on the 2-colored and the 3-colored … So

the number of positions reachable by just the 2-colored [and

3-colored] pieces is (24!x32!)/2"

He did mention that neither 2- or 3-colored stickers could switch just

two pieces, but he only divided by two once, not compensating for both

occurrences. The real value, I believe, should have been half of what

he gave us in that part of the equation.