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An employer has 6 applicants for a programming position
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10 Sep 2016, 07:10

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85% (01:14) correct 15% (01:34) wrong based on 101 sessions

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An employer has 6 applicants for a programming position and 4 applicants for a manager position. If the employer must hire 3 programmers and 2 managers, what is the total number of ways the employer can make the selection?

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10 Sep 2016, 07:21

azamaka wrote:

An employer has 6 applicants for a programming position and 4 applicants for a manager position. If the employer must hire 3 programmers and 2 managers, what is the total number of ways the employer can make the selection?

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19 Nov 2017, 14:49

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azamaka wrote:

An employer has 6 applicants for a programming position and 4 applicants for a manager position. If the employer must hire 3 programmers and 2 managers, what is the total number of ways the employer can make the selection?

a) 1,490 b) 132 c) 120 d) 60 e) 23

Take the task of selecting employees and break it into stages.

Stage 1: Select 3 programmers to hire Since the order in which we select the programmers does not matter, we can use combinations. We can select 3 programmers from 6 programmers in 6C3 ways (20 ways) So, we can complete stage 1 in 20 ways

Stage 2: Select 2 managers to hire Once again, we can use combinations We can select 2 managers from 4 applicants in 4C2 ways (6 ways) So, we can complete stage 2 in 6 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 (and thus hire all of the people) in (20)(6) ways (= 120 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

Re: An employer has 6 applicants for a programming position
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22 Nov 2017, 12:25

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azamaka wrote:

An employer has 6 applicants for a programming position and 4 applicants for a manager position. If the employer must hire 3 programmers and 2 managers, what is the total number of ways the employer can make the selection?

a) 1,490 b) 132 c) 120 d) 60 e) 23

The programmers can be selected in 6C3 = 6!/3![(6-3)!] = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2) = 20 ways.

The managers can be selected in 4C2 = 4!/[2!(4-2)!] = (4 x 3)/2! = 6 ways.

Thus, the total number of ways to select the group is 20 x 6 = 120.

Answer: C
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