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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

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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