Number/Color CubeDate: 09/13/2001 at 07:50:45 From: bill welden Subject: Number/color cube You have a cube that you want to make into a number cube by putting the numbers 1,2,3,4,5,6 on the faces. The only restriction is that you put 1 and 5, 3 and 6, 2 and 4 on opposite faces. Each face is a different color (red, orange, yellow, green, blue, purple). How many ways can you make the cube? Answer I'm trying to solve: 6 sides to a cube, 6 different colors, 3 different number combinations, 2 different numbers in each pair. Date: 09/13/2001 at 11:22:35 From: Doctor Mitteldorf Subject: Re: Number/color cube Dear Bill, There may be powerful, general methods for attacking this problem, but I'm going to think about it from the ground up, in a very concrete way. First, about the colors: how many ways are there to put colors on the faces? Answer: red can be paired opposite 5 different colors. Once you've decided which color is opposite red, there are 4 colors left, which can be paired on opposite faces in 3 different ways (e.g., {yellow- green, and blue-purple}, {yellow-blue, and green-purple}, {yellow- purple, and blue-green}). So 5*3 makes 15 distinct ways to assign pairs of colors to the pairs of faces, and there's one more freedom that we have: the mirror-image of each of these paintings is a distinct way in itself. For example, suppose the pairing is {red-orange, yellow-green, blue- purple}. Then there's one corner that has red, yellow, and blue faces coming together. Looking down on that corner, do those colors appear (red,yellow,blue) clockwise or (red, yellow, blue) counter-clockwise? These 2 mirror-image possibilities, multiplied by the 15 pairings, indicates a total of 30 ways to paint the cube. Second, once we've chosen a paint scheme, how many ways are there to assign the numbers, consistent with the pairings (1-5, 3-6, 2-4)? Again, we'll subdivide this question into two parts: first, how to match the number-pairs with the color-pairs; second, how to choose which number goes with which color within those pairings. So, the first part of the second part: Each of these 3 number pairs can be matched with any of the 3 color pairs. The number of ways of doing this is the same as the number of ways of ordering 3 objects, which is 3!=6. This is because you can regard the color-pairings as a threesome in fixed order, and then each ordering of the number-pairs corresponds to a way to match them against these. And for the final part of the second part: Once you've decided which of the 30 colorings to use, and you've decided which of the 6 color- pair, number-pair matchings to use, you still have a choice within each pair. For each color-number pair (e.g. red-orange labeled 1-2), you can choose one of two labelings: (e.g. either 1 is red and 2 is orange, or 1 is orange and 2 is red). You have 3 such choices to make, and they are independent, so there are 2^3=8 ways of doing the assignment. Putting all this together, I get 5*3*2 * 3! * 2^3 = 1440 different dice. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/