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

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

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19 Jun 2008, 08:17
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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
Intern
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19 Jun 2008, 08:21
I just listed out the combinations:

Office 1 : Office 2
A : BC
B : AC
C : AB
AB : C
AC : B
BC : A
ABC : {null}
{null} : ABC

Manager
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19 Jun 2008, 08:40
i did the same but it is too long...
I think there must be fast aproach here!
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19 Jun 2008, 08:46
Yeah I was thinking the same thing, but I think combinatorics at this level of number of options would be overkill and would take more time than just listing them out. The problem is that you have to arrange all options for each office 2(3C0 + 3C1 + 3C2 + 3C3) THEN you have to subtract all the options that overlap. That just boggles the mind.

Maybe there is an easy combination that I am missing for this one.
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19 Jun 2008, 08:50
1
KUDOS
Each employee can be assigned in 2 ways - hence in total it is 2x2x2 = 8
Manager
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19 Jun 2008, 08:52
looks like a good aproach here
kudo!
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19 Jun 2008, 08:55
1
KUDOS
this is a labelling problem
you can assign each employee with Office 1 or Office 2.
so that would make 2!
2! * 2! * 2! = 8
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19 Jun 2008, 09:07
1
KUDOS
The more I thought about it all you have to do is know a bit about set theory and find out all the possible sets with three objects. i.e.

number of objects: number in set

1 : 2
2 : 2
3 : 8
4 : 16

So the 2x2x2 is a product of set theory as well. Set cardinality = 2^(# of objects).
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19 Jun 2008, 09:18
+1 for you

Wouldn't that only work for 2 sets?
What if you had 8 employees and 5 offices?
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19 Jun 2008, 09:26
Yep you are right. Because we know that figuring out one offices population is all we need with two offices we can use set theory. With 5 offices we have multiple possibilities where a office would be empty. So knowing the sets for one is not enough. Hmm, now I want to figure out that one
Manager
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19 Jun 2008, 09:32
jmaynardj
In this case I suppose it must be 56
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19 Jun 2008, 09:48
jmaynardj wrote:
I just listed out the combinations:

Office 1 : Office 2
A : BC
B : AC
C : AB
AB : C
AC : B
BC : A
ABC : {null}
{null} : ABC

This basically was my approach.

ABC 0 <- 2 ways
AB C <- 2ways
AC B <- 2ways
BC A <- 2 ways

8 ways.

D
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19 Jun 2008, 09:56
Ok this took some back figuring, but I believe a rule for this type of problem holds as follows:

2 offices 3 employees : 2^3
5 offices 8 employees : 5^8 = 390625
So take n office with x employees : n^x

This is just an extension off of a pattern I observed... I was not able to come up with a proof (especially at work ) But what is funny is that the rule turns out to be "Each employee can be assigned in 2 ways - hence in total it is 2x2x2 = 8" which is what iamcartic said. I am not sure how to get at it from set theory, but it does show that set theory and combinatorics are not so dissimilar.
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19 Jun 2008, 10:31
Hmm yah I think that works out, I don't have the proof either...@ work
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19 Jun 2008, 11:54
How about 5 offices, 8 employees and one office can not have more than 4 employees?
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19 Jun 2008, 13:08
Expert's post
rajesh04 wrote:
How about 5 offices, 8 employees and one office can not have more than 4 employees?

This is for case if only one office cannot have more than 4 employees.

$$N=4^8+C^8_1*4^7+C^8_2*4^6+C^8_3*4^5+C^8_4*4^4$$
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19 Jun 2008, 13:18
Can you explain especially 5C1 part...
Thanks!!
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19 Jun 2008, 13:27
Expert's post
rajesh04 wrote:
Can you explain especially 5C1 part...
Thanks!!

It was for other problem (if only one office can have more than 4 employees) and I deleted this part of the post
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19 Jun 2008, 13:30
5C1 => 5 Combination 1 => 5!/(1!(5-1)!) = 5

nCr = n!/(r!(n-r)!)
Re: PS combinatory   [#permalink] 19 Jun 2008, 13:30
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