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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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    Your answer in the case where order doesn't matter is incorrect. If you have 10 balls satisfying the condition, we have a constraint on 6 of them leaving 4 remaining balls with complete freedom. I make it 10.2018-12-08

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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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    I'll post separate thread then. Or you can expand your answer refering comments below.2011-11-10