# In how many ways 3 flags of colors black, purple & yellow can be arranged at the corners of an equilateral triangle?

In how many ways 3 flags of colors black, purple & yellow can be arranged at the corners of an equilateral triangle?

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Welcome to MathSE. I see that this is your first question. So I wanted to let you know a few things about MathSE. We like to know the sources of questions - if it's homework, please add the [homework] tag. People will still help, so don't worry. We also like to know what you've tried on a problem or what your thoughts are, so that the answer does not re-invent the wheel. These sort of pleasantries usually result in more and better answers. Thank you! –  Arturo Magidin Nov 23 '11 at 16:38
Hint: you are finding the number of $ordered$ arrangements of three $distinct$ objects. –  David Mitra Nov 23 '11 at 16:55
I can see a considerable number of possible interpretations of this question. For a start, it could be three distinguisable flags (one black, one purple, and one yellow) or three indistinguishable tricolour flags. Then there may or may not be an implied constraint that each corner has precisely one flag positioned at it. Finally the corners or the triangle may or may not be distinguishable. –  Peter Taylor Nov 23 '11 at 19:26

There is only a very finite amount of possibilities... Try to find out the distinct cases:

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Those don't look like equilateral triangles :) –  David Mitra Nov 23 '11 at 17:16
@DavidMitra due to my perfect programming skills it has to be an optical illusion :) –  Listing Nov 23 '11 at 17:18
@DavidMitra: If "Geometry is making correct deductions about incorrectly drawn figures", then maybe combinatorics is "correctly counting incorrectly listed possibilities." (-: –  Arturo Magidin Nov 23 '11 at 17:25

This appears to me as a problem of circular permutation.

Fix one flag of any color (say black) at a particular corner of the specified equilateral triangle, then the other two flags (purple and yellow) can be arranged in $2!$ ways at the other two corners. Hence, the answer should be $4$.

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2! isn't 4. The answer is either 2 or 6 depending on what the question is (which isn't clear). –  Peter Taylor Nov 23 '11 at 19:23