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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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11 Nov 2008, 19:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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11 Nov 2008, 19:48
gorden 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

4 ways to fill in one office => (0,3) (1,2) (2,1), (3,0)
2 such offices
total ways = 2*4 = 8

D
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12 Nov 2008, 02:22
alpha_plus_gamma wrote:
4 ways to fill in one office => (0,3) (1,2) (2,1), (3,0)
2 such offices
total ways = 2*4 = 8
D

Can you elaborate the highlighted steps further? Is (0,3) (1,2) (2,1), (3,0) not taking care of both the offices?
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12 Nov 2008, 03:06
It is not mentioned that every employee has to be employed in one of the offices.

Hence the QA could be E , thats is 9 ways.
0,3
1,2
2,1
3,0
0,1
0,2
1,0
2,0
1,1

If this is not true i cannot see more than 4 ways as stated by scthakur, but thats not in the answer choices. Hence it could be the above method.
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12 Nov 2008, 03:16
1
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Expert's post
The problem says: " In how many ways can the company assign 3 employees to 2 different offices?"
So, all 3 employees are assigned to one of two offices.

There two possibilities for each employee: first office or second office. So, N=2^3=8 (D)
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12 Nov 2008, 05:18
That makes sense, I don't know why i was'nt thinking in that way. Something happens to me when i see probability ...
Whats the QA btw ?
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13 Nov 2008, 04:03
walker wrote:
The problem says: " In how many ways can the company assign 3 employees to 2 different offices?"
So, all 3 employees are assigned to one of two offices.

There two possibilities for each employee: first office or second office. So, N=2^3=8 (D)

thanks walker for an excellent explanation.
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13 Nov 2008, 13:02
walker wrote:
The problem says: " In how many ways can the company assign 3 employees to 2 different offices?"
So, all 3 employees are assigned to one of two offices.

There two possibilities for each employee: first office or second office. So, N=2^3=8 (D)

Hey Walker

you mean this right

3C1+3C2+3C3=2^3=8
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13 Nov 2008, 14:01
hibloom wrote:

Hey Walker

you mean this right

3C1+3C2+3C3=2^3=8

3C1+3C2+3C3=3+3+1=7
Anyway, I don't see logic behind your formula....Maybe you mean 3C0+3C1+3C2+3C3? If yes, I think it is right.
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14 Nov 2008, 14:08
sorry that was a mistake
i meant
3c0+3c1+3c2+3c3=8

Thanks
Re: company--27   [#permalink] 14 Nov 2008, 14:08
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