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A certain company employs 6 senior officers and 4 junior of

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ericnkem
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A certain company employs 6 senior officers and 4 junior of [#permalink] New post   29 Nov 2010, 15:33
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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 committee are possible?

A. 8
B. 24
C. 58
D. 80
E. 210
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D
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Bunuel
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Re: Combinations!!! [#permalink] New post   29 Nov 2010, 15:40
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ericnkem 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 committee are possible?

a- 8
b-24
c- 58
d-80
e-210


\(C^3_6*C^1_4=80\): \(C^3_6\) - # of ways to choose 3 different senior officers out of 6 and \(C^1_4\) - # of ways to choose 1 junior officer out of 4. We multiply these two because if one event can occur in \(m\) ways and a second can occur independently of the first in \(n\) ways, then the two events can occur in \(mn\) ways (this is called Principle of Multiplication).


Answer: D.

For theory on combinatorics check: math-combinatorics-87345.html

Hope it helps.
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Re: A certain company employs 6 senior officers and 4 junior of [#permalink] New post   30 Aug 2016, 17:38
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ericnkem 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 committee 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 ways

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 ways

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

Answer: D
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gettinit
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Re: Combinations!!! [#permalink] New post   29 Nov 2010, 22:50
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6!/3!(6-3)! = 20

4!/1!(4-1)!=4

so number of different committees equals 20*4=80
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Re: A certain company employs 6 senior officers and 4 junior of [#permalink] New post   03 Sep 2017, 10:54
ericnkem 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 committee are possible?

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



Basically in this question we need to select 3 senior officer from the 6 senior officers available and 1 junior officer from 4 junior officers available.

This can be done in 6C3 * 4C1 ways
= [6*5*4/(3*2)]*[4C1]
= 20*4
= 80

Answer D
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A certain company employs 6 senior officers and 4 junior of [#permalink] New post   03 Sep 2017, 11:02
There are 6 senior officers and 4 junior officers.
The committee contains 3 senior officers and 1 junior officer.

3 senior officers can be selected from 6 officers in 6C3 ways = (6 x 5 x 4)/3! = 20 ways
1 junior officers can be selected from 4 officers in 4c1 ways= 4 ways

Therefore total number of ways to select 3 senior and 1 junior officer are 20*4 =80 ways.
Hence Answer is D.

Kudos if it helps. :-)
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Re: A certain company employs 6 senior officers and 4 junior of [#permalink] New post   02 Jul 2018, 06:09
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ericnkem 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 committee 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. So, be sure to learn it.

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Re: A certain company employs 6 senior officers and 4 junior of [#permalink] New post   02 Jul 2018, 10:01
The order is not important hence this is a combinations questions.

The number ways in which a committee can be made with 3 senior officers and 1 junior officer in each committee is in 6C3 * 4C1 = 80 ways.

Option D is the correct answer
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Re: A certain company employs 6 senior officers and 4 junior of [#permalink] New post   03 Jun 2019, 06:53
ericnkem 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 committee are possible?

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


combination
6c3*4c1 = 20*4 ; 80
IMO D
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Re: A certain company employs 6 senior officers and 4 junior of [#permalink] New post   03 Nov 2023, 04:38
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Re: A certain company employs 6 senior officers and 4 junior of [#permalink]
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