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We have 3 different GREEN balls, 4 different RED balls and 2 different YELLOW balls. Find the number of ways in which the balls can be chose so that atleast 1 GREEN ball and 1 YELLOW ball are chosen.

P.S: Note that we can select any number of balls. It can be 1 ball to all 9 balls.

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There are $2^9$ subsets of the set of $9$ balls. We will count the bad subsets, the subsets that contain no green or no yellow.

There are $2^6$ subsets with no green, and $2^7$ with no yellow. The sum $2^6+2^7$ double-counts the subsets with no green and no yellow. There are $2^4$ of these.

Thus the number of bad subsets is $2^6+2^7-2^4$. The number of subsets with at least one green and at least one yellow is therefore $2^9-2^6-2^7+2^4$.

Remark: We have used the Principle of Inclusion/Exclusion.

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  • $\begingroup$ Thanks for the note on the principle used $\endgroup$ May 21 '14 at 15:55
  • $\begingroup$ You are welcome. $\endgroup$ May 21 '14 at 16:38
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Ways to choose at least 1 GREEN $\times$ Ways to choose at least 0 RED $\times$ ways to choose at least 1 YELLOW.

$$\left(\binom31+\binom32+\binom33\right)\times\left(1+\binom41+\binom42+\binom43+\binom44\right)\times\left(\binom21+\binom22\right)$$

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  • $\begingroup$ which is $2^4(2^2 - 1)(2^3-1) =$ the other answer. $\endgroup$
    – PA6OTA
    May 19 '14 at 22:08
  • $\begingroup$ Thank you for providing an alternate way to solve it. $\endgroup$ May 21 '14 at 15:55

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