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An enterprise has five departments. There are three managers in each

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An enterprise has five departments. There are three managers in each  [#permalink]

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New post 10 Dec 2017, 10:36
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A
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  55% (hard)

Question Stats:

69% (01:59) correct 31% (02:38) wrong based on 54 sessions

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An enterprise has five departments. There are three managers in each of the five departments. A committee of four has to be made such that not more than one manager comes from each department. What is the total of the different committees possible?

A. 60
B. 81
C. 255
D. 405
E. 455

Source: Experts Global

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Re: An enterprise has five departments. There are three managers in each  [#permalink]

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New post 10 Dec 2017, 14:38
15*12*9*6/5! = 81 —> B


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Re: An enterprise has five departments. There are three managers in each  [#permalink]

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New post 10 Dec 2017, 15:40
Since maximum of 1 manager can be chosen from each department. The question comes down to:
1. Selecting 4 dept out of 5
2. Select 1 manager out of 3 in each of the departments chosen in (1) above.

Ways to choose 4 departments = 5
Ways to choose 1 manager out fo 3 = 3

Since there are a total of 4 departments, the total possibilities = 5x3x3x3x3 = 405
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Re: An enterprise has five departments. There are three managers in each  [#permalink]

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New post 10 Dec 2017, 20:26
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pushpitkc wrote:
An enterprise has five departments. There are three managers in each of the five departments. A committee of four has to be made such that not more than one manager comes from each department. What is the total of the different committees possible?

A. 60
B. 81
C. 255
D. 405
E. 455

Source: Experts Global


\(C^4_5*3^4=405\):

5C4 is the number of ways to choose which 4 departments out of 5 will delegate managers to the committee. Each of those 4 chosen departments can send any of the three members it has, so 3*3*3*3 = 3^4.

Answer: D.
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Re: An enterprise has five departments. There are three managers in each  [#permalink]

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New post 15 Dec 2017, 09:01
I answered this type of question correctly for first time! (Go me).

The way I think about it is:
Number of options for first manager = 15
for second manager= 12
for third= 9
fourth= 6.
But 15*12*9*6 is permutation and as we don't care about order here since we are picking teams (can pick teams in any order, the teams will remain the same unless specified about order).

Therefore we divide by 4! as 4 managers can be arranged in 4! way. Answer will be =(15*12*9*6)/4! = 405.
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Re: An enterprise has five departments. There are three managers in each &nbs [#permalink] 15 Dec 2017, 09:01
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