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Monday, August 24, 2015

Problem of the Week

Summer as we know it is almost over! If you're not trying to avoid anything academic for as long as you can, here's this week's Problem of the Week.


I've posted the solution after the break, so you don't have to break out your buckets of paint unless you really want to.




Did you figure out the solution in a different way? What about the number of ways to color other polyhedra? Let us know in the comments!


2 comments:

  1. There are 6 faces, so there are 6! = 6x5x4x3x2x1 choices to paint the cube if the cube is fixed. But the order of symmetry group of cube is 4! = 4x3x2x1. Thus the answer is 6! / 4! = 6x5 = 30.

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    Replies
    1. There are 6 faces, so there are 6! = 6x5x4x3x2x1 choices to paint the cube if the cube is fixed.
      But the 6 faces can be the floor face : 6!/6=120.
      While a face is on the floor, 4 positions are possible. 120/4=30.
      Same answer, not (?) same reasons. I wonder if my solution is good ... (not sure)

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