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by yvonne12 » Wed Mar 07, 2007 8:54 am
acertain 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 selections does not matter?

ans.3060

please explain
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by ssiva » Wed Mar 07, 2007 10:24 am
The formula is (n!)/[(n-r)! * r!] for a combination problem

18!/[(14!)(4!)] = (18*17*16*15)/(1*2*3*4) = 3060
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by Jeff@TargetTestPrep » Sun Jun 17, 2018 7:12 pm
yvonne12 wrote:acertain 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 selections does not matter?

ans.3060

please explain
Since order does not matter, 4 people can be chosen from 18 in:

18C4 = 18!/(4! x 14!) = (18 x 17 x 16 x 15)/4! = (18 x 17 x 16 x 15)/(4 x 3 x 2) = 3 x 17 x 4 x 15 = 3,060 ways.

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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