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# How many committees can be formed comprising 2 male members selected f

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8033
GMAT 1: 760 Q51 V42
GPA: 3.82
How many committees can be formed comprising 2 male members selected f  [#permalink]

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04 Mar 2019, 02:02
00:00

Difficulty:

15% (low)

Question Stats:

82% (01:21) correct 18% (02:21) wrong based on 33 sessions

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[GMAT math practice question]

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900
B. 1200
C. 1500
D. 1800
E. 2400

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 25 Feb 2019 Posts: 336 Re: How many committees can be formed comprising 2 male members selected f [#permalink] ### Show Tags 04 Mar 2019, 02:30 imo B . 2 men of 4 , 3 women out of 5 and 3 juniors out of 6 4C2*5C3*6C3 = 1200 Posted from my mobile device Intern Joined: 21 Apr 2014 Posts: 32 Re: How many committees can be formed comprising 2 male members selected f [#permalink] ### Show Tags 04 Mar 2019, 04:05 We could almost rephrase this as three different questions: How many ways to choose 2 from 4? How many ways to choose 3 from 5? How many ways to choose 3 from 6? For this question, it's easier if you know the Combinations formula, but you could technically count. Combinations formula = n! / k! (n-k)! n = the bigger number (what we're choosing from) k = the smaller number (how many we're choosing) For example: how many ways to choose 2 from 4? 4! / 2! (4-2)! 4! / 2!2! (4 x 3 x 2 x 1) / (2 x 1)(2 x 1) 24/4 = 6 But if we counted this and said the four males were ABCD, we could just list AB, AC, AD, BC, BD, and CD, and you can see it also equals 6. Anyway, finding the other two: How many ways to choose 3 from 5? 5! / 3! (5-3)! 5! / 3! 2! We can cancel out 3! from both numerator and denominator. 5 x 4 / 2 20/2 = 10 How many ways to choose 3 from 6? 6! / 3! (6-3)! 6! / 3! 3! We can cancel out one 3! from both numerator and denominator. 6 x 5 x 4 / 3 x 2 120 / 6 = 20 Back to our original questions: How many ways to choose 2 from 4? 6 How many ways to choose 3 from 5? 10 How many ways to choose 3 from 6? 20 Now, multiply all those numbers together! 6 x 10 x 20 = 1200 _________________ Vivian Kerr | GMAT/GRE Tutor @ http://www.gmatrockstar.com | gmatrockstar[at]gmail.com | solid 5-star reviews on Yelp! Former Kaplan and Grockit instructor. I've freelanced with Veritas Prep, Magoosh, and most of the bigger test prep companies. Now offering Skype-based private tutoring for the GMAT and GRE. Reading Comprehension is my jam! Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8033 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: How many committees can be formed comprising 2 male members selected f [#permalink] ### Show Tags 06 Mar 2019, 02:02 => There are 4C2 ways to select $$2$$ men from $$4$$ men, 5C3 ways to select $$3$$ women from $$5$$ women and 6C3 ways to select $$3$$ juniors from $$6$$ juniors. Therefore, the total number of possible committees is 4C2*5C3*6C3 = $${\frac{(4*3)}{(2*1)}}{\frac{(5*4*3)}{(3*2*1)}}{\frac{(6*5*4)}{(3*2*1)}} = 6*10*20 = 1200.$$ Therefore, B is the answer. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: How many committees can be formed comprising 2 male members selected f  [#permalink]

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06 Mar 2019, 02:11
MathRevolution wrote:
[GMAT math practice question]

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900
B. 1200
C. 1500
D. 1800
E. 2400

4c2 * 5c3*6c3 = 1200
IMO B
Director
Joined: 27 May 2012
Posts: 905
Re: How many committees can be formed comprising 2 male members selected f  [#permalink]

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09 Mar 2019, 04:30
1
MathRevolution wrote:
[GMAT math practice question]

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900
B. 1200
C. 1500
D. 1800
E. 2400

4c2 * 5c3*6c3 = 1200

P.S. Question should be tagged Combinations rather than probability.
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- Stne
Re: How many committees can be formed comprising 2 male members selected f   [#permalink] 09 Mar 2019, 04:30
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