To fill a number of vacancies, an employer must hire 3

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

To fill a number of vacancies, an employer must hire 3

by AAPL » Wed Aug 29, 2018 4:42 am

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Princeton Review

To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection?

A. 1,490
B. 132
C. 120
D. 60
E. 23

OA C.

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Source: Beat The GMAT — Problem Solving | Wed Aug 29, 2018 4:42 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Wed Aug 29, 2018 5:01 am

AAPL wrote:Princeton Review

To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection?

A. 1,490
B. 132
C. 120
D. 60
E. 23

Number of ways to choose 3 programmers from 6 applicants = 6C3 = (6*5*4)/(3*2*1) = 20.
Number of ways to choose 2 managers from 4 applicants = 4C2 = (4*3)/(2*1) = 6.
To combine the options above, we multiply:
20*6 = 120.

The correct answer is C.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Wed Sep 05, 2018 9:43 am

AAPL wrote:Princeton Review

To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection?

A. 1,490
B. 132
C. 120
D. 60
E. 23

This is a combination problem since the order of selecting the applicants for each job title does not matter.

3 programmers from 6 applicants can be hired in 6C3 ways:

6C3 = 6!/3!(6-3)! = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2 x 1) = 20

2 managers from 4 applicants can be hired in 4C2 ways:

4C2 = 4!/2!(4-2)! = (4 x 3)/2! = (4 x 3)/(2 x 1) = 6

Thus, the total number of ways in which she can make her selection is 20 x 6 = 120.

Answer: C

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