Find the number of distinguishable ways the edges of a square can be painted if six colors of paint are available and the same color may be used on more than one edge.

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- November 18th 2011, 05:34 PMveronicak5678Group Motions
Find the number of distinguishable ways the edges of a square can be painted if six colors of paint are available and the same color may be used on more than one edge.

- November 18th 2011, 06:03 PMDevenoRe: Group Motions
try to break down:

1) the number of ways the square's edges can be painted with k colors, for k = 1,2,3,4. how does knowing the symmetries of the square help here?

2) the number of ways we can choose k colors from 6. - November 18th 2011, 06:25 PMveronicak5678Re: Group Motions
I'm confused about why this isn't jut a combinatorics problem.

1) 6 choices for each edge = 6*4

2)6 choose k = 6!/ ((6-k)!k!) - November 18th 2011, 07:07 PMDevenoRe: Group Motions
your answer for (1) is incorrect.

suppose we have just one color. then all ways of painting the square's edges are indistinguishable.

suppose we have 2 colors. there are 6 different ways to choose 2 edges out of 4. this gives 6 different ways to paint the square with any 2 given colors.

another way to see this, is that we can either a) paint two adjacent sides, or b) paint two opposite sides. adjacency is preseved under rotation,

but the vertices are not. this gives us 4 ways (one for each rotation) to paint adjacent sides. rotation also preserves opposite sides, but "rotation squared"

gives us a square indistinguishable from our original square, so we only get 2 ways to paint opposite sides.

suppose we have 3 colors. one color gets painted to two sides, there are 6 different ways for this to happen, and 3 colors for it to happen with.

for each of those 18 ways, there are 2 ways to choose to paint the remaining 2 sides, giving 36 ways to paint the square with 3 colors (and we haven't even chosen the colors yet!).

finally, we have 4! ways to paint the square with 4 colors: 4 choices for the 1st color, having done that 3 choices for the 2nd color, and 2 choices for the 3rd color. - November 18th 2011, 07:19 PMveronicak5678Re: Group Motions
Ohhh... I didn't realize what they meant by distinguishable. Thanks!

- November 28th 2011, 05:59 PMveronicak5678Re: Group Motions
Does this seem right? I am using the Burnside theorem:

r= 1/|G| * sum of |distinct orbits|

so I have |G|=8, |S1|= 6^4, |Sx|=6, |Sx^2|=6^2, |Sx^3|=6, |Sy|=|SXy|=|Sx^2y|=|Sx^3y|=6^2

so r = (1/8)(1488)=186