Number of possibilities, triangles with different colors?

Hi everybody,

 

I'm doing research with triangles. Here's my problem:

 

Imagine three points (a triangle). There are four possible colors to color an edge with. You can use the colors as often as you'd like. How many different triangles can you form? The answer isn't as straightforward as 4^3, because a rotation is not a "new triangle". My answer by just writing all 4^3=64 possibilities down and the categorize them gave 16 different possibilities. Would any of you know how to theoretically get this number instead of just writing them all down? The attachement shows all 16 triangles (in my case i swapped the four colors for four different arrows, but that doesn't matter)

 

Secondly, would anyone know a formula to count these possibilities? That if you would add a point that you'll be able to calculate the amount of possibilities (If you'd add a point you'll get a pyramid). My guess at first was 4^(n-1) with n being the amount of points. But if you would get a lot of points I don't think it doesn't take into account that the 3d models become symmetric, so it would count too much possibilities.

 

Thanks!

RattenG

Motifs_10.pdf

Hi Rattan and welcome to the forum

 

For a start I think I get 24 not 16 - I haven't drawn them but have a look at this

 

1) Four Colours ABCD

 

2) Equilateral triangle so ABC = BCA = CAB but does NOT equal ACB = CBA = BAC

 

3) So we can divide differently coloured triangles as follows

 

a. monochrome - one colour

 

AAA BBB CCC DDD

 

b. 2+1 - two sides similar colour

 

2*A => AAB AAC AAD

2*B => BBA BBC BBD

2*C => CCA CCB CCD

2*D => DDA DDB DDC

 

c. 1+1+1 - all three different ie missing one colour

no A => BCD BDC

no B => ACD ADC

no C => ABD ADB

no D => ABC ACB

 

a = 4

b = 3 * 4

c = 2 * 4

 

Total = 24

 

 

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For equilateral triangles:

 

'a' would always equal number of colours

 

'b' would always equal number of colours x (number of colours minus 1)

 

'c' is more complicated and would be [number of colours CHOOSE (number - 3)] x 2

 

NB

n CHOOSE k or [latex]{n \choose k}[/latex] is the combination (wikipage here) and evaluates at [latex]\frac{n!}{k!(n-k)!}[/latex]

 

For a square - next most complicated, I will think if there is an easy progression. the third dimension really makes things tough

 

 

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for a square

 

1, monochrome -> Number of Colours (from here called C). No patterns; ie all arrangments are identical under rotation

 

2. 3 same plus one different -> C x (C-1) . No patterns; ie all arrangments are identical under rotation

 

3. 2 same plus 2 -> (C x (C-1)) / 2. Multiplied by 2 patterns for each colour set ie abab aabb

 

4. 2 same plus 1 plus 1 -> (C x (C-1) x (C-2))/2. Multiplied by 3 patterns for each colour set ie aabc aacb abac

 

5. All different [latex]\frac{C!}{C!(C-4)!}[/latex] Multiplied by 6 patterns for each colour set ie abcd abdc acbd acdb adbc adcb

 

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for a tetrahedron (6 edges)

 

there are 11(this strikes me a wrong) ways to pick the colours - and each way has a certain number of unique patterns those colours can be arranged in. but after 3 one colour and 3 a different colour - my ability to visualize rotations gives out and i would have to be really formal, and I wont. there are 12 orientations of a tetrahedron - so that's got to come into a generalization but I cannot for the life of me see one at present.

Why equilateral?

Why equilateral?

 

I used equilateral triangle because the OP stated

 

because a rotation is not a "new triangle"

 

In an isoceles triangle or an irregular triangle you can define each side uniquely - rotating the finished object does not make one triangle appear like another . For Isoceles you would have short side, first long side clockwise from shortside, and second longside clockwise from shortside. For irregular you would have shortist, middle and longest side. If you can define each side uniquely ABC does NOT equal CAB.

 

But for an equilateral triangle you can just rotate a triangle coloured ABC by 180 degrees and you, in effect, have one coloured CAB. ie ABC = CAB = BCA

 

Not equilateral would make it easier - as you would have no repetitions to remove.

So why don't you do it the simple way and count/remove the repetitions?

Edited January 4, 201312 yr by michel123456

So why don't you do it the simple way and count/remove the repetitions?

 

Because the repetitions vary on the different schemes. If you know a simple way - post it. There is surely an easy way to do it - but I cannot see it. It seems clear that it should be analogous to choosing 3 objects from a set of 4 - but I cannot yet systematize (ie know without thinking a/o checking) which patterns are unique and which are just rotations of another.

A rectangle equilateral is the same under 3 rotations and under mirroring.

what does that give mathematically I don't know.