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# If 1 male, 2 females, and 1 child are to be randomly selected from 8 m

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
GMAT 1: 760 Q51 V42
GPA: 3.82
If 1 male, 2 females, and 1 child are to be randomly selected from 8 m  [#permalink]

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13 Feb 2017, 00:00
1
00:00

Difficulty:

15% (low)

Question Stats:

83% (01:29) correct 17% (02:13) wrong based on 38 sessions

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If 1 male, 2 females, and 1 child are to be randomly selected from 8 males, 10 females, and 8 children, how many such cases are possible?

A. 980
B. 1,440
C. 1,880
D. 2,480
E. 2,880

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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 $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 17 May 2015 Posts: 249 Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m [#permalink] ### Show Tags 13 Feb 2017, 00:38 MathRevolution wrote: If 1 male, 2 females, and 1 child are to be randomly selected from 8 males, 10 females, and 8 children, how many such cases are possible? A. 980 B. 1,440 C. 1,880 D. 2,480 E. 2,880 Hi, Few Important points: AND => Multiplication OR => Addition. When Order is not important => Apply Combination formula When Order is important => Apply Permutation formula. If 1 male, 2 females, and 1 child are to be randomly selected from 8 males, 10 females, and 8 children, how many such cases are possible? In the above problem order is not important, so we'll use Combination formula. We have to select 1 male from 8 males, AND 2 females from 10 females, AND 1 child from 8 children. = $$^{8}C_1 \times ^{10}C_{2} \times ^{8}C_{1} = 8 \times 45 \times 8 = 2880$$ Thanks. Current Student Joined: 27 May 2014 Posts: 525 GMAT 1: 730 Q49 V41 Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m [#permalink] ### Show Tags 13 Feb 2017, 01:16 1 male can be chosen from 8 males in 8 ways; 1 child can be chosen from 8 children in 8 ways; 2 females can chosen from 10 females: 1. 1st female can be chosen in 10 ways; There are 9 females remaining; 2. so, 2nd female can be chosen in 9 ways; Now the order does not matter; so, total possible selection for 2 females (10*9)/2! = 45; Therefore, total possible cases = 8*8*45 = 2880; Answer E. _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6815 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m [#permalink] ### Show Tags 15 Feb 2017, 00:08 ==> You get 8C1*10C2*8C1=(8)(45)(8)=2,880. The answer is E. Answer: E _________________ 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$149 for 3 month Online Course"
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Re: If 1 male, 2 females, and 1 child are to be randomly selected from 8 m &nbs [#permalink] 15 Feb 2017, 00:08
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