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Number/Color Cube


Date: 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/   
    
Associated Topics:
College Discrete Math
High School Discrete Mathematics
High School Permutations and Combinations

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