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Painted Cube Faces


Date: 03/22/2001 at 15:12:25
From: Paul
Subject: Combination problem

Each face of a cube, numbered 1 through 6, is to be painted either 
red or blue. How many different possible combinations are there, 
from all blue faces to all red faces and every combination in between?


Date: 03/22/2001 at 15:28:10
From: Doctor Anthony
Subject: Re: Combination problem

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 three 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, and 
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 w = white as the two colours.)

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

=(1/24)[24w^6 +24r^6 +24rw^5 +24r^5w +48r^2w^4 +48r^3w^3 +48r^4w^2]

and putting r = 1, w = 1 this gives

   240/24  = 10 non-equivalent configurations.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 03/26/2001 at 12:57:13
From: Doctor Greenie
Subject: Re: Combination problem

Hello, Paul -

Excuse me for stepping in where another doctor provided the initial 
response to your question.

The wording of your original question allows different interpretations 
of the problem. Dr. Anthony's answer of 10 is correct for one of the 
interpretations.

One interpretation is that you are merely looking for the number of 
different ways to paint the six faces either red or blue. In this 
interpretation, if the cube is a standard die with the faces numbered 
from 1 to 6, then (for example) each of the six cases with a single 
face red and the others blue is a "different" way of painting the cube 
- because red 1, red 2, red 3, red 4, red 5, and red 6 are 
distinguishable. 

In the other interpretation, you are looking for distiguishably 
different ways of painting the faces of the cube. In this 
interpretation, the faces of the cube are not labeled, so all six 
cases of a single face red with the others blue look the same; they 
are indistiguishable from each other.

With the first interpretation of the problem, the number of ways to 
paint the cube is simply 2*2*2*2*2*2 = 64, because you have two 
choices for the color for each face.

With the second interpretation of the problem, Dr. Anthony's answer is 
correct. I am not familiar with the mathematical methods he used to 
come up with his answer; however, this problem is easy enough to 
visualize, and a little analysis shows 10 distinguishably different 
ways to paint the faces either red or blue.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

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