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

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

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04 Feb 2005, 23:14
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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
Senior Manager
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05 Feb 2005, 00:36
HongHu wrote:
2*2*2=8?

yaa man

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05 Feb 2005, 09:55
hmmm....why not 3*3 = 9 ?? Each office can contain all 3 employees ??
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05 Feb 2005, 11:14
Look at it this way:
First employee can be assigned to either 1st or 2nd office (thus we have 2 options here)
Second employee can be assigned to either 1st or 2nd office (thus 2 more options)
Third employee can be assigned to either 1st or 2nd office (2 more options)
Thus, we have 2*2*2=8
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05 Feb 2005, 14:06
Marina wrote:
Look at it this way:
First employee can be assigned to either 1st or 2nd office (thus we have 2 options here)
Second employee can be assigned to either 1st or 2nd office (thus 2 more options)
Third employee can be assigned to either 1st or 2nd office (2 more options)
Thus, we have 2*2*2=8

Yep I see that...but what if the ques said , how many ways 2 offices can be assigned to 3 different employees ?? I think it will be 3*3 in that case, won't it ?
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05 Feb 2005, 22:20
3*3 would mean that the first office gets to be assigned to one of the three people, then the second office gets to be assigned to one of the three people again. In other words, each office can have only one person (meaning one person is without an office), and it is possible that both office is assigned to one person (meaning two people are without an office).

If it is two reports that are to be assigned to three people to type, and it is possible for a person to type two reports, than the answer would be 3*3.
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05 Feb 2005, 23:17
I'm also getting 8 as the answer. Here is my reasoning:

There are 2 offices and 3 people, and each person needs to be assigned to an office. There are 4 ways this could happen: (3,0), (2,1), (1,2), (0,3). In other words, you could have all 3 assigned to the 1st office, 2 assigned to the 1st office and 1 assigned to the 2nd office, etc. Each of the cases where all 3 are in the same office can only be done in 1 way. Each of the cases with 2 in 1 office and 1 in the other can be done in (3c2) or 3 ways. So you have 1+3+3+1=8 total ways.

I know its a lot more complicated than the other way, but it is what I thought of when I saw the problem and I solved it pretty quickly.
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06 Feb 2005, 01:17
GOOD JOBS GUYS

TO ME, I THINK 2*2*2 is the most easy way to solve it
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07 Feb 2005, 11:30
I agree with Banerjeee

Each office can have all 3 employees.

but the question says some of the offices can have 0 employees... thus the answer will be 3*3 - 1 = 8

DLMD can u post the OE.
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07 Feb 2005, 13:54
OK this was rather easy as number of employees and offices were small numbers.
If you have E employees and O offices where does it lead ??
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07 Feb 2005, 15:37
The Correct answer is 2^3 = 8.

The no. of ways = no of ways each employee can be assigned a office

1st employee can be assigned in 2 ways (Office 1 or 2)
2nd employee can be assigned in 2 ways (Office 1 or 2)
3rd employee can be assigned in 2 ways (Office 1 or 2)

Multiply , you get 8.

There is a formula for distributing n things among r partitions (where each partition can be empty) = r^n.

For our problem n = 3 and r = 2.
2^3=8   [#permalink] 07 Feb 2005, 15:37
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