Probabiltiy Q help

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sunil_snath: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Wed Nov 18, 2009 8:41 am; Thanked: 3 times

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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Source: Beat The GMAT — Problem Solving | Sun Jan 10, 2010 7:42 am

rohan_vus: Master | Next Rank: 500 Posts; Posts: 189; Joined: Thu Apr 03, 2008 2:03 pm; Location: USA; Thanked: 21 times

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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