Rubik's Cube Comb

For discussing Olympiad Level Combinatorics problems
Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

Rubik's Cube Comb

A $3 \times 3$ Rubik's cube has different colors on its each sides.
In how many different ways the cube can be arranged ?
Name: Fahim Shahriar Shakkhor
Notre Dame College

nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

Re: Rubik's Cube Comb

see the permutations part.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

Re: Rubik's Cube Comb

I'm saying what I did.

The centers are fixed. Excluding centers there 20 small cubes outside. 8 of them are in the corners and 12 in middles.

12 middles can be arranged in 12! ways and as each middles has two colors, permutation for middles = $12!\times 2$

8 corners can be arranged in 8! ways and each has 3 colors. (According to rules they should be arranged in $3!=6$ ways. But actually they can be in 3 ways. Do it practically.) So permutations of corners = $8!\times 3$

TOTAL = $12!\times 2\times 8!\times 3$
Last edited by Phlembac Adib Hasan on Mon Jan 28, 2013 3:01 pm, edited 1 time in total.
Reason: use \times instead of x(to show a $\times$)
Name: Fahim Shahriar Shakkhor
Notre Dame College

sakibtanvir
Posts: 188
Joined: Mon Jan 09, 2012 6:52 pm
Location: 24.4333°N 90.7833°E

Re: Rubik's Cube Comb

How?? Shouldn't it be $12! \times 2^{12} \times 8! \times 3^{8} ???$
I don't know If I am right or wrong...For the corner pieces the approach should be as below:
"There are $8$ corner pieces,Each piece having $3$ colors.For our advantage,at first we assume that each corner piece is painted with a distinct color(Not with the combination of 3).The permutation will be $8!$.Now we consider their position at this time we assume the corner pieces painted with 3 colors(their original state!).Name the Pieces as below:$A,B,C,D,E,F,G,H$.Notice,$A$ can hold 3 positions,$B$ can hold 3 positions,.....,H can hold 3 positions.So the total number of arrangement for the corner pieces will be $8! \times 3^{8}$.I have spent those word just to clarify my idea.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

Fahim Shahriar
Posts: 138
Joined: Sun Dec 18, 2011 12:53 pm

Re: Rubik's Cube Comb

@sakibtanvir. You're right. I did the permutation for only one piece and forgot to do for others. Thanks to correct me..

So finally we get
$12! \times 2^{12} \times 8! \times 3^{8}$
Name: Fahim Shahriar Shakkhor
Notre Dame College