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GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
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Oct 17 · Chris Peckover
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Oct 20 · Chris Peckover
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Oct 25 · Josh Braslow
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Sep 6 to Dec 6, 2026

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40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
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by sunil_snath » Sun Jan 10, 2010 7:42 am
Given 5 different green dyes, 4 different blue dyes, 3 different red dyes, how many combination of dyes can be chosen taking atleast 1 green and 1 blue dye?

A. 4096
B. 3255
C. 3720
D. None of these.

Ans: [spoiler] C

[/spoiler]
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by rohan_vus » Sun Jan 10, 2010 8:12 am
One dye can be chosen as nC0 + nC1...nCn = 2^n..But this slection incudes a case of nC0 where the dye was not chosen..

So if you need atleast 1 dye in the slection than that dye can be selected as 2^n - 1 ( 1 yuu deduct because you are kicking out the case of nC0)...

So Green can be selected in 2^5 -1 = 31
Blue can be selected in 2^4 -1 = 15
Red can be 2^3 = 8

So ans = 31*15*8 = 3720
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