A certain company has 18 equally qualified applicants for 4

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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

The OA is E

Source: GMAT Prep

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by GMATGuruNY » Mon Jan 28, 2019 8:05 am
swerve 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
From 18 applicants, the number of ways to choose 4 = 18C4 = (18*17*16*15)/(4*3*2*1) = (3*17*4*15) = 51*60 = 3060.

The correct answer is E.
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by [email protected] » Mon Jan 28, 2019 10:09 am
Hi All,

We're told that a certain company has 18 equally qualified applicants for 4 open positions. We're asked for the number of different groups of 4 applicants that can be chosen by the company to fill the positions if the order of selection does not matter. This question is a fairly straight-forward Combination Formula question (Mitch's solution shows how to work through that formula). Once you recognize that that formula can be applied, you can actually avoid doing some of that math though:

The numerator of that Combination Formula calculation will include the number 17. Since that number is a prime number, there's nothing in the denominator that can 'reduce it' - meaning that the correct answer MUST be a multiple of 17. As such, you can quickly eliminate the first 3 answers (since they are clearly NOT multiples of 17). With a little work, you can also eliminate Answer D (since it's not a multiple of 17 either). That just leaves the correct answer...

Final Answer: E

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by Scott@TargetTestPrep » Wed Jan 30, 2019 6:00 pm
swerve 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

The OA is E

Source: GMAT Prep
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.

Answer: E

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