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How many combinations of three colors using one alone, two together, or three together?

I get 36, but I don't know the formula, or if I miscounted!

R Y B RR YY BB RY YR BR RB YB BY RYB YRB BRY RBY YBR BYR RRY YYR BBR RRB YYB BBY RBR YRY BRB RYR YBY BYB YRR RYY RBB BRR BYY YBB

Thanks, Michael

Michael Lester
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    You missed BBB, RRR and YYY. Also, if you look at just 1, just 2 and just 3 separately, I think you will see a pattern emerge in those numbers. – Arthur Jun 19 '19 at 07:39
  • Thank you, Arthur! Yes, I did miss three of them! – Michael Lester Jun 19 '19 at 07:42
  • It's easier to get them all if you're being systematic, following as simple a pattern as possible. Write down all the ones that start with R first, then all the ones that start with Y, then all the ones that start with B. Now, within each of those, first write down the ones which have R in the middle, then the ones that have Y in the middle, then the ones that have B in the middle. This should really point you straight towards why the given answer below is correct. – Arthur Jun 19 '19 at 07:44
  • Thank you, I was trying to be systematic, but I think I petered out... I started with all the single colors, then all the two colors, then all the three, but I did miss three, like you said. Thanks, again! – Michael Lester Jun 19 '19 at 07:57
  • I was thinking specifically about the three-colour combinations. They look chaotic and it's no wonder you missed a few. I see you first tried all combinations with all three colours, then all combinations that use one colour twice, but 1) that's still a bit chaotic for my taste, and 2) it doesn't make it easy to find a solution. – Arthur Jun 19 '19 at 07:59
  • Chaos is the natural order of things in my muddled mind, I think! – Michael Lester Jun 19 '19 at 08:00

1 Answers1

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You can choose a single color in $3$ different ways. Having a sequence of $n$ colors, the amount of different choices is $3\times3\times\cdots\times3=3^n$

Thus, in your case:
$3^1+3^2+3^3=39$

In general case, sequences lengths from $1$ to $n$ with $k$ choices, one would have $$\sum_{j=1}^nk^j=\frac{k^{n+1}-1}{k-1}-1$$ different choices.

M.P
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  • Thank you very much, M.P.! It's great to know the formula. So, 39 is correct! – Michael Lester Jun 19 '19 at 07:44
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    So, here is my application. I have to select 25 poems from about 60 poets who may submit up to five poems each. I will judge blind and have an associate print one poem on one index card, which could amount to 300, plus poems. My associate will use color codes on the index card to show which poems belong to one particular poet. 39 color combinations is not enough, but I think 64 will work. So, using the formula you showed me, I would use four colors instead of three, and allow any combination of three of the four colors. 4 to the 3rd power. – Michael Lester Jun 19 '19 at 07:53