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There are three bags with red balls, blue balls and green balls. Each of the bags contains at least 10 balls of a single color.

How many ways you can choose exactly 10 balls containing at least one red ball, at least two green balls and at least three blue balls?

Now, you can understand this question two ways. If order of picking balls matters or not.

When not, there is a $6\choose2$ = 15 possible ways how to choose. It is easy. But I can't figure out, how many when order matters? How to achieve it? I asked my math teacher but he doesn't know.

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1 Answer

up vote 1 down vote accepted

You can use inclusion-exclusion.

There are $3^{10}$ ways to pick 10 balls if the order does matter.

If at least one of the conditions fails, then:

  • Not enough red balls; there are $2^{10}$ ways of picking with no red balls;
  • Not enough green balls; $2^{10}$ ways with no green balls, $10\times 2^9$ with exactly one green ball;
  • Not enough blue balls; $2^{10}$ with no blue balls, $10\times 2^9$ with exactly one, $\binom{10}{2}\times 2^8$ with exactly two.

If at least two of the conditions fail, then

  • Not enough red and green balls: $1$ ways with no red or green balls at all; $10$ with no red and exactly one green ball.
  • Not enough red and blue balls; $1$ way with no red or blue balls at all; $10$ with no red and exactly one blue; $\binom{10}{2}$ with no red and exactly two blue.
  • Not enough green and blue; $1$ way with no green or blue; $10$ with no green and exactly one blue; $\binom{10}{2}$ with no green and exactly two blue; $10$ with exactly one green and no blue; $10\times 9$ ways with exactly one green and one blue; $10\times\binom{9}{2}$ with exactly one green and two blue.

You cannot have all three conditions fail.

So in summary, we have: $$\begin{align*} 3^{10} &- (2^{10} + 2^{10}+10\times 2^9 + 2^{10}+10\times 2^9+45\times 2^{8}) \\ &+ (1 + 10 + 1 + 10 + 45 + 1 + 10 + 45 + 10 + 90 + 360)\\ &= 34800. \end{align*}$$

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Awesome. No chance I can figure it out by my self. –  Ondrej Janacek Nov 9 '11 at 21:44
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@Andrew: Why not? It's simply counting the different possibilities... –  Arturo Magidin Nov 9 '11 at 21:57
    
Because I thought about it as a universal problem. What if you have more colors? It isn't practical to work with more and more conditions. It solved my problem, but still I am interested whether there is a shorter universal method. –  Ondrej Janacek Nov 10 '11 at 7:55
    
@Andrew: Then may I suggest that you either unaccept my answer, edit your question to add that you would like a more general approach that might work for $k$ balls of $m$ different colors, with at least $a_1$ of color $1$, $a_2$ of color $2$, ..., and $a_m$ of color $m$, etc; or else post it as a separate question refering to this one? –  Arturo Magidin Nov 10 '11 at 14:24
    
I'll post separate thread then. Or you can expand your answer refering comments below. –  Ondrej Janacek Nov 10 '11 at 20:56
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