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# a07a - CRN 87401 MAD 6206 Combinatorics I Fall 2011...

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MAD 6206 Combinatorics I, Fall 2011 CRN: 87401 Assignment 7 – suggested solution Question 1 . (Rotations of the 3-d cube) We talked about coloring the 6 faces of the cube in class so that each face has a different color. We shall color the vertices and edges of the cube in this problem. Assume that colorings are equivalent under rotations. (a) Color the eight vertices of the cube with k colors so that no two vertices have the same color. How many different colorings are there? (b) Color the twelve edges of the cube with k colors so that no two edges have the same color. How many different colorings are there? Solution . Let G be the group containing the 24 rotations of the cube. (a) Using Burnside’s Lemma, the number of different colorings of the vertices of the cube with k colors so that no two vertices have the same color equals 1 24 summationdisplay g G ψ ( g ) . Since the eight vertices have different colors, the only rotation that fixes a coloring is the identity permu- tation ι and ι fixes ( k ) 8 colorings. Hence, the answer is 1 24 ( k ) 8 . (b) From Burnside’s Lemma again, the number of different colorings equals 1 24 summationdisplay g G ψ ( g ) . Since all edges have different colors, the only rotation that fixes a coloring is the identity which fixes all ( k ) 12 colorings. Thus, the answer is 1 24 ( k ) 12 . Question 2 . (Cycle index of the group of rotations on the vertices of the cube) Consider coloring the faces of the cube, assuming equivalence under rotations. Let G be the corresponding permutation group on the set of vertices of the cube. (a) Find the cycle index of G . (b) Find the number of possible colorings if there are m colors available. (c) Find the number of possible colorings if every one of the m available colors is used. Solution . (a) There are 24 permutations. (A) For the identity, there are 8 cycles of length 1. Hence contribution is X 8 1 . (B) For the three axes through the centers of opposite faces, a π/ 2 rotation has 2 cycles with length 4; a π rotation has 4 cycles of length 2; a 3 π/ 2 rotation has the same cycles as a π/ 2 rotation. Hence contribution for this case: 3 ( 2 X 2 4 + X 4 2 ) .

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a07a - CRN 87401 MAD 6206 Combinatorics I Fall 2011...

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