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# A certain company assigns employees to offices in such a way

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A certain company assigns employees to offices in such a way [#permalink]

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20 Sep 2008, 07:20
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A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9
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20 Sep 2008, 07:26
i think it is 3P2 = 6

But when i try to enumerate I am able to think of following cases only, Am I missing something ?

03
12
21
30
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20 Sep 2008, 14:57
8

each employee has 2 ways to go
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20 Sep 2008, 15:11
i get 9..

3C1+3C2+2C1+1C1=9
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20 Sep 2008, 21:12
amitdgr wrote:
i think it is 3P2 = 6

But when i try to enumerate I am able to think of following cases only, Am I missing something ?

03
12
21
30

Say a,b and c are the 3 employees

I think 8.

-----ROOM 1 ----|---- ROOM2-----

-----NONE-------|-----abc---------
-----abc---------|-----NONE---------
-----a-----------|------bc---------
-----b-----------|------ac---------
-----c-----------|------ab---------
-----bc----------|------a---------
-----ac----------|------b---------
-----ab----------|------c---------

did i miss any case ?
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22 Sep 2008, 11:37
Does the question also imply that none of the employees can be assigned to any of the two offices?
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22 Sep 2008, 23:46
Nihit wrote:
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

scthakur wrote:
Does the question also imply that none of the employees can be assigned to any of the two offices?

The question states "some of the offices can be empty"
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23 Sep 2008, 07:46
i just realized 8 is correct..each employee has 2 possibilites...2^3
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23 Sep 2008, 07:59
fresinha12 wrote:
i just realized 8 is correct..each employee has 2 possibilites...2^3

Can you explain this please ??
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23 Sep 2008, 08:08
amitdgr wrote:
fresinha12 wrote:
i just realized 8 is correct..each employee has 2 possibilites...2^3

Can you explain this please ??

Employe 1 can be placed in two ways (either office A or B)
Employe 2 can be placed in two ways (either office A or B)
Employe 3 can be placed in two ways (either office A or B)

No of ways all three can be assigned. = 2*2*2=8
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23 Sep 2008, 12:16
x2suresh wrote:
amitdgr wrote:
fresinha12 wrote:
i just realized 8 is correct..each employee has 2 possibilites...2^3

Can you explain this please ??

Employe 1 can be placed in two ways (either office A or B)
Employe 2 can be placed in two ways (either office A or B)
Employe 3 can be placed in two ways (either office A or B)

No of ways all three can be assigned. = 2*2*2=8

Thanks Suresh makes a lot of sense now. Cheers to you buddy
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Re: Employees   [#permalink] 23 Sep 2008, 12:16
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