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An employer has 6 applicants for a programming...

by swerve » Sun Nov 19, 2017 1:40 pm
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 OA is C.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.
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by Brent@GMATPrepNow » Sun Nov 19, 2017 1:48 pm
swerve 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

If anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
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MEDIUM
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- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Jeff@TargetTestPrep » Wed Nov 22, 2017 11:26 am
swerve 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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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