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A container has $20$ black balls, $15$ white balls and $18$ red balls. Each $2$ balls of the same color are identical. How many ways there are to extract balls from the container such that we choose at least $1$ black ball, $1$ white ball and $1$ red ball. No other limitations on the amount of balls extracted apply.

What I tried to do is apply inclusion/exclusion principle and count the number of possibilities where:

  • $A_1$ = No black ball was extracted
  • $A_2$ = No white ball was extracted
  • $A_3$ = No red ball was extracted

Because each ball is either extracted or not, the total number of possibilities is $2^{53}.$ Applying the inclusion/exclusion we result in: $$|A_1| + |A_2| + |A_3| = 2^{33} + 2^{38} + 2^{35}$$ $$|A_1 \cap A_2| + |A_1 \cap A_3| + |A_2 \cap A_3| = 2^{15} + 2^{18} + 2^{20}$$ $$|A_1 \cap A_2 \cap A_3| = 1$$

The total number of possibilities is: $2^{53} - 2^{33} - 2^{38} - 2^{35} + 2^{15} + 2^{18} + 2^{20} - 1$

However the textbook's answer is: $20 \cdot 18 \cdot 15$. Just by looking at this, I have no idea how one can come up with an answer like this. Can someone please explain what am I doing wrong?

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  • $\begingroup$ I think you make things more difficult than necessary. Concerning the black balls there are $20$ choices for the number of balls to be extracted: $1,2,3,4,\dots,20$. Likewise there are $18$ choices for the red balls and $15$ for the white balls. That gives $20\times18\times15$ choices in total. $\endgroup$ – drhab Jan 23 '18 at 9:01
  • $\begingroup$ Thanks, great explanation. Consider re-writing this as an answer so I could accept it. :) $\endgroup$ – 0rka Jan 23 '18 at 9:03
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You are making things more difficult than necessary.

Concerning the black balls there are $20$ choices for the number of balls to be extracted: $1,2,3,4,\dots,20$.

Likewise there are $18$ choices for the red balls and $15$ for the white balls.

That gives $20\times18\times15$ choices in total.

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