 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  FIGURE it out (2) (Posted on 2004-09-25) 1. How many different tetrahedrons can be produced by coloring each face a solid color and using n different colors? (Two tetrahedrons are the same if they can be turned and placed side by side so that corresponding sides match in color.)

2. How many cubes with n colors?

 No Solution Yet Submitted by SilverKnight Rating: 4.5000 (4 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution-part 1 | Comment 2 of 14 | I'm going to take a similar approach to the one I used on figure it out 1.  This one is different, because there are much more ways you can rotate the solids, and I'm much more likely to miss one.
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Tetrahedrons:

First, there is a total of n^4 tetrahedrons, but many of them are simply rotations of each other.  For unique coloration, I think there are 12 total rotations.  My reasoning is that if you track one of the four faces, it can go in any of the four positions, and in each position it can be rotated in three ways.  4 times 3 is 12.  So, the n^4 tetrahedrons would mostly come in groups of 12 rotations.

However, some don't come in groups of 12.  If all 4 faces are colored the same way, then that coloration only rotates to itself.  Also, if 3 are colored the same, and the 4th is different, then it would only have 4 rotations.  If there are two pairs of different sides, then there are 6 rotations.

To sum up the above two paragraphs, there are 4 categories of tetrahedron color schemes:
1) those that come in groups of 1
2)...groups of 4
3)...groups of 6
4)...groups of 12

There are n type 1 tetrahedrons.  Self-explanatory, I hope.

There are n*(n-1) groups of type 2 tetrahedrons.  This is because the three sides can be any of the n colors, and the last side must be any of the n-1 different sides.  Note that I said groups of tetrahedrons.  When I count n^4 total tetrahedrons, I count each single group as four.

There are n*(n-1)/2 groups of type 3 tetrahedrons.  This is because the color combinations of the two pairs of faces is C(2,n)=n*(n-1)/2.

All the rest of the n^4 tetrahedrons are type 4.  That would be n^4-n-4n(n-1)-6n(n-1)/2 tetrahedrons (simplify to n(n+3)(n-2)(n-1).  That means there are n(n+3)(n-2)(n-1)/12 groups of type 4 tetrahedrons.

The total of all that is this:
n^2(n^2+11)/12

I'd like to write the whole solution for both parts in the same comment, but I'm afraid the comment will get deleted or something, so I'll come up with the second part soon enough.

 Posted by Tristan on 2004-09-25 16:49:07 Please log in:
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