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Colour Combinations on a Cube

Date: 02/08/2002 at 06:48:00
From: Daniel
Subject: Number of colour combinations on a cube

If each side of a cube is painted red or yellow or blue, how many 
distinct colour patterns are possible? So far I haven't managed to 
solve any part of this problem.

Date: 02/08/2002 at 07:27:26
From: Doctor Anthony
Subject: Re: Number of colour combinations on a cube

This is a problem requiring the use of Burnside's lemma. You consider 
the 24 symmetries of a cube and sum all those symmetries that keep 
colours fixed. The number of non-equivalent configurations is then the 
total sum divided by the order of the group (24 in this case).

We first find the cycle index of the group of FACE permutations 
induced by the rotational symmetries of the cube.

Looking down on the cube, label the top face 1, the bottom face 2 and 
the side faces 3, 4, 5, 6 (clockwise)

You should hold a cube and follow the way the cycle index is 
calculated as described below. The notation 

   (1)(23)(456) = (x1)(x2)(x3) 

means that we have a permutation of 3 disjoint cycles in which face 1 
remains fixed, face 2 moves to face 3 and 3 moves to face 2, face 4 
moves to 5, 5 moves to 6 and 6 moves to 4. (This is is not a possible 
permutation for our cube, it is just to illustrate the notation.) We 
now calculate the cycle index.

(1)e = (1)(2)(3)(4)(5)(6);  index = (x1)^6
(2)3 permutations like (1)(2)(35)(46); index 3(x1)^2.(x2)^2
(3)3 permutations like (1)(2)(3456); index 3(x1)^2.(x4)
(4)3 further as above but counterclockwise; index 3(x1)^2.(x4) 
(5)6 permutations like (15)(23)(46); index 6(x2)^3
(6)4 permutations like (154)(236); net index 4(x3)^2
(7)4 further as above but counterclockwise; net index 4(x3)^2

Then the cycle index is

P[x1,x2,...x6] =(1/24)[x1^6 + 3x1^2.x2^2 + 6x2^3 + 6x1^2.x4 + 8x3^2]

and the pattern inventory for these configurations is given by the 
generating function:  

(I shall use r=red and b=blue,  y=yellow as the three colours)

f(r,b,y) = (1/24)[(r+b+y)^6 + 3(r+b+y)^2.(r^2+b^2+y^2)^2 
           + 6(r^2+b^2+y^2)^3 + 6(r+b+y)^2.(r^4+b^4+y^4) 
           + 8(r^3+b^3+y^3)^2]

and putting r=1, b=1, y=1 this gives

  =(1/24)[3^6 + 3(3^2)(3^2) + 6(3^3) + 6(3^2)(3) + 8(3^2)]

  = (1/24)[729 + 243 + 162 + 162 + 72]

  = 1368/24

  = 57   

So there are 57 non-equivalent configurations.

- Doctor Anthony, The Math Forum
Associated Topics:
High School Permutations and Combinations
High School Symmetry/Tessellations

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