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A certain company has 18 equally qualified applicants for 4 [#permalink]
07 Mar 2008, 03:12
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Question Stats:
87% (01:48) correct
13% (01:07) wrong based on 247 sessions
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
A certain company has 18 equally qualified applicants for 4 [#permalink]
07 Sep 2010, 09:47
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
Re: GMATPrep 18 equally qualified applicants for 4 open position [#permalink]
07 Sep 2010, 09:59
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Expert's post
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jpr200012 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
(A) 18 (B) 72 (C) 180 (D) 1,260 (E) 3,060
# of ways to choose 4 different people out of 18, when order of chosen people doesn't matter is \(C^4_{18}=\frac{18!}{14!*4!}=3060\).
Re: A certain company has 18 equally qualified applicants for 4 [#permalink]
02 May 2014, 10:44
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Re: A certain company has 18 equally qualified applicants for 4 [#permalink]
06 Jun 2015, 23:31
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A certain company has 18 equally qualified applicants for 4 [#permalink]
09 Aug 2015, 05:19
Expert's post
japped187 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
(A) 18 (B) 72 (C) 180 (D) 1,260 (E) 3,060
Initially There are 18 choices for first Position Now, There are 17 choices for Second Position Now, There are 16 choices for Third Position Now, There are 15 choices for Forth Position
i.e. Total Ways to choose people (With arrangement) = 18*17*16*15
Since we require only the selection hence we need to exclude the arrangements of 4 selected individuals which is 4!
i.e. i.e. Total Ways to choose people (WithOUT arrangement) = (18*17*16*15)/4! = 3060
Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!
Re: A certain company has 18 equally qualified applicants for 4 [#permalink]
22 Oct 2015, 12:51
Bunuel wrote:
jpr200012 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
(A) 18 (B) 72 (C) 180 (D) 1,260 (E) 3,060
# of ways to choose 4 different people out of 18, when order of chosen people doesn't matter is \(C^4_{18}=\frac{18!}{14!*4!}=3060\).
A certain company has 18 equally qualified applicants for 4 [#permalink]
22 Oct 2015, 13:00
1
This post received KUDOS
GMATDemiGod wrote:
Bunuel wrote:
jpr200012 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
(A) 18 (B) 72 (C) 180 (D) 1,260 (E) 3,060
# of ways to choose 4 different people out of 18, when order of chosen people doesn't matter is \(C^4_{18}=\frac{18!}{14!*4!}=3060\).
Answer: E.
if order mattered how would this problem change?
If the order mattered, then the 4 people you selected out of 18 via 18C4 need to be multiplied by 4 to account for the fact that 4 positions will themselves be arranged in 4! ways.
Thus total ways possible with an ordered set = 18C4* 4! _________________
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