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A certain company employs 6 senior officers and 4 junior

Expert replies
by M7MBA » Mon Jul 02, 2018 1:41 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created, that is made up of 3 senior officers and 1 junior officer, how many different committees are possible?

A. 8
B. 24
C. 58
D. 80
E. 210

The OA is D.

What is the formula that I should use here? Please, I need some help here. Thanks in advance.
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Source: — Problem Solving |

by GMATGuruNY » Mon Jul 02, 2018 2:18 am
M7MBA wrote:A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created, that is made up of 3 senior officers and 1 junior officer, how many different committees are possible?

A. 8
B. 24
C. 58
D. 80
E. 210
From 6 senior officers, the number of ways to choose 3 for the committee = 6C3 = (6*5*4)/(3*2*1) = 20.
Since there are 4 junior officers, the number of options for the junior officer on the committee = 4.
To combine the options above, we multiply:
20*4 = 80.

The correct answer is D.
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by Shahrukh_mbabreakspace » Mon Jul 02, 2018 2:33 am
We have selected 3 people out of 6 and simultaneously 1 out of 4:
So, we have to do 6C3*4C1= 80
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by Brent@GMATPrepNow » Mon Jul 02, 2018 5:02 am
M7MBA wrote:A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created, that is made up of 3 senior officers and 1 junior officer, how many different committees are possible?

A. 8
B. 24
C. 58
D. 80
E. 210
Take the task of creating the 4-person committee and break it into stages.

Stage 1: Select 3 senior officers
Since the order of the selected officers does not matter, we can use combinations.
We can select 3 officers from 6 senior officers 6C3 ways (= 20 ways)

Stage 2: Select 1 junior officer
There are 4 junior officers, so we can select 1 officer in 4 ways

By the Fundamental Counting Principle (FCP) we can complete the two stages (and thus select members for the committee) in (20)(4) ways (= 80 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: 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
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Wed Jul 04, 2018 6:14 pm
M7MBA wrote:A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created, that is made up of 3 senior officers and 1 junior officer, how many different committees are possible?

A. 8
B. 24
C. 58
D. 80
E. 210
We need to determine how many different committees are possible with 3 senior officers and 1 junior officer from 6 senior officers and 4 junior officers. Let's first determine the number of ways to select 3 senior officers.

Number of ways to select 3 senior officers from 6 of them = 6C3 = (6 x 5 x 4)/3! = 20

Next we can determine the number of ways to select 1 junior officer.

Number of ways to select 1 junior officer from 4 of them = 4C1 = 4

Thus, the number of ways to select 3 senior officers and 1 junior officer is 20 x 4 = 80.

Answer: D

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