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A certain company employs 6 senior officers and 4 junior of

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A certain company employs 6 senior officers and 4 junior of [#permalink]

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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
[Reveal] Spoiler: OA

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Re: Combinations!!! [#permalink]

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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: Combinations!!! [#permalink]

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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]

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Re: A certain company employs 6 senior officers and 4 junior of [#permalink]

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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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Re: A certain company employs 6 senior officers and 4 junior of [#permalink]

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Re: A certain company employs 6 senior officers and 4 junior of [#permalink]

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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]

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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.

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A certain company employs 6 senior officers and 4 junior of   [#permalink] 03 Sep 2017, 11:02
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