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/
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