tag:blogger.com,1999:blog-5159574672638793285.post4115070775301501410..comments2024-01-29T06:01:15.858-05:00Comments on The Center of Math Blog: Problem of the WeekCenter of Mathematicshttp://www.blogger.com/profile/06584574038886207370noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5159574672638793285.post-67917755095132260062015-08-24T13:23:40.141-04:002015-08-24T13:23:40.141-04:00There are 6 faces, so there are 6! = 6x5x4x3x2x1 c...There are 6 faces, so there are 6! = 6x5x4x3x2x1 choices to paint the cube if the cube is fixed.<br />But the 6 faces can be the floor face : 6!/6=120.<br />While a face is on the floor, 4 positions are possible. 120/4=30.<br />Same answer, not (?) same reasons. I wonder if my solution is good ... (not sure)maelnoreply@blogger.comtag:blogger.com,1999:blog-5159574672638793285.post-78953729835154886962015-08-24T09:36:58.998-04:002015-08-24T09:36:58.998-04:00There are 6 faces, so there are 6! = 6x5x4x3x2x1 c...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.Sooji Shinhttp://soojishin.comnoreply@blogger.com