Distinguishing tetrahedrons with colors

[V0ODO0CH1LD]

Homework Statement

How many regular tetrahedrons (of the same dimension) can you distinguish using 4 different colors to paint all of its 4 faces?

Homework Equations

The Attempt at a Solution

I got to 36, but I am not confident about my answer because I used mostly "real world" intuition. How can I model this mathematically?

If I used one color each time I could only paint the whole tetrahedron using that one color. So that's 4 different tetrahedrons right there, one of each different color.

If I used two different colors I could paint one face using the first color and still have 3 options for the remaining 3 faces. So that's 4*(3), 12 more options.

I could also paint two faces using one color and the other two using another. But then I would get some similar tetrahedrons by simple rotation.

How can I model this??

 

[verty]

Look up permutations, this question is one of counting permutations.

 

[haruspex]

How can I model this mathematically?

Permutation questions where you have to eliminate duplicates that arise through symmetries are tough to solve in a generic way. For this question, I would be content to go through the different cases, as you have.

I could also paint two faces using one color and the other two using another. But then I would get some similar tetrahedrons by simple rotation.

So how many are you counting for this case?
How about 3 different colors? 4 different (this one is a bit tricky)?

 

[V0ODO0CH1LD]

Permutation questions where you have to eliminate duplicates that arise through symmetries are tough to solve in a generic way. For this question, I would be content to go through the different cases, as you have.

So how many are you counting for this case?
How about 3 different colors? 4 different (this one is a bit tricky)?

It's really tricky, and the worst part is that I don't know how to check my answer. I am almost building a tetrahedron to see for myself. Is it me or are these problems a pain to model?

I think I did the right thing by separating the different cases into one color, two colors, three colors and four colors. But I don't know how to account for the symmetries.

 

[skiller]

I'm probably being a bit thick here, but if you have 4 faces and 4 colours, then fix the base of the tetrahedron to one colour, then you can only paint the other three faces in a certain number of distinct ways.

So how many combinations in total does that give you? Seems simple to me, but I'm not telling...

 

[haruspex]

I don't know how to check my answer.

Post your attempts and I will check them.

Is it me or are these problems a pain to model?

can't say I find it that hard. Depends how well you can visualise objects.

I think I did the right thing by separating the different cases into one color, two colors, three colors and four colors. But I don't know how to account for the symmetries.

You did fine so far. Given two colors, how many ways of painting two faces in each color?

 

[haruspex]

fix the base of the tetrahedron to one colour, then you can only paint the other three faces in a certain number of distinct ways.

That is not going to work. If you reuse the base color on another face then you may find you've counted the same arrangement twice.

 

[V0ODO0CH1LD]

Here is how I did it:

one color (4 options)
- the whole tetrahedron painted with each color

two colors (12 + 6)
- twelve coming from painting one face with one color, and the remaining three with another one of the three remaining colors.
- again I have four options for the first two faces and three for the remaining two, but this time I have to account for symmetries by simple rotation. And I figured half of them would be the same as the other half.

three colors (12)
- in this case I have four options for the first face, three for the second, and two options for the remaining two faces. Except that again half are the same. So that is only 12.

four colors (2)
- there are only two arrangements that are not symmetric

total = 36 options

 

[verty]

Sorry, I misinterpreted the question. I read it as asking not "how many are there" but "how many can you distinguish", as in, if it looks different, it is different. I thought it was simply 4^4 paintings for this reason. Although why I thought that is beyond me.

I know what happened. You know how sometimes you think to yourself "I mustn't drop this" and just then you drop it? Something like that happened. I remember I was thinking I must keep the post quality high, being a homework helper. I was thinking that just before reading this first post.

 

[verty]

@V0ODO0CH1LD: that looks fine. These questions tend to be about feel and intuition, imagining a real situation. I don't think there is a satisfactory answer to your question about whether there is a general mathematical approach to this. I mean, apart from the usual summing of cases like you did.

 

[haruspex]

Here is how I did it:

one color (4 options)
- the whole tetrahedron painted with each color

two colors (12 + 6)
- twelve coming from painting one face with one color, and the remaining three with another one of the three remaining colors.
- again I have four options for the first two faces and three for the remaining two, but this time I have to account for symmetries by simple rotation. And I figured half of them would be the same as the other half.

three colors (12)
- in this case I have four options for the first face, three for the second, and two options for the remaining two faces. Except that again half are the same. So that is only 12.

four colors (2)
- there are only two arrangements that are not symmetric

total = 36 options

That's all the same as I got.
In principle, the question can be posed more abstractly by specifying the symmetry group ("A4" in this case) instead of the object it comes from. There might then be formulae that could be applied relating the numbers of colourings to the subgroups (see http://en.wikipedia.org/wiki/Tetrahedral_symmetry#Chiral_tetrahedral_symmetry), but if so it's beyond my knowledge, and would anyway be inappropriately sophisticated for the present question.

 

[skiller]

That is not going to work. If you reuse the base color on another face then you may find you've counted the same arrangement twice.

My mistake.

I thought it was 4 distinct colours - one on each face - ie you couldn't use the same colour on more than one face.

If I'd actually read the posts preceding mine, I would've realized that. D'oh! Sorry...