A certain company assigns employees to offices in such a way : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 18 Jan 2017, 12:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 14 May 2008
Posts: 70
Followers: 1

Kudos [?]: 88 [0], given: 0

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

### Show Tags

19 Jun 2008, 07:17
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

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
Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

Kudos [?]: 8 [0], given: 0

### Show Tags

19 Jun 2008, 07:21
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
Joined: 14 May 2008
Posts: 70
Followers: 1

Kudos [?]: 88 [0], given: 0

### Show Tags

19 Jun 2008, 07:40
i did the same but it is too long...
I think there must be fast aproach here!
Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

Kudos [?]: 8 [0], given: 0

### Show Tags

19 Jun 2008, 07:46
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.
Manager
Joined: 24 Apr 2008
Posts: 162
Followers: 1

Kudos [?]: 43 [1] , given: 0

### Show Tags

19 Jun 2008, 07:50
1
KUDOS
Each employee can be assigned in 2 ways - hence in total it is 2x2x2 = 8
Manager
Joined: 14 May 2008
Posts: 70
Followers: 1

Kudos [?]: 88 [0], given: 0

### Show Tags

19 Jun 2008, 07:52
looks like a good aproach here
kudo!
Intern
Joined: 05 Nov 2007
Posts: 35
Followers: 0

Kudos [?]: 2 [1] , given: 0

### Show Tags

19 Jun 2008, 07:55
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
Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

Kudos [?]: 8 [1] , given: 0

### Show Tags

19 Jun 2008, 08:07
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).
Intern
Joined: 05 Nov 2007
Posts: 35
Followers: 0

Kudos [?]: 2 [0], given: 0

### Show Tags

19 Jun 2008, 08:18
+1 for you

Wouldn't that only work for 2 sets?
What if you had 8 employees and 5 offices?
Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

Kudos [?]: 8 [0], given: 0

### Show Tags

19 Jun 2008, 08:26
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
Joined: 14 May 2008
Posts: 70
Followers: 1

Kudos [?]: 88 [0], given: 0

### Show Tags

19 Jun 2008, 08:32
jmaynardj
In this case I suppose it must be 56
CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19

Kudos [?]: 421 [0], given: 0

### Show Tags

19 Jun 2008, 08:48
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
Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

Kudos [?]: 8 [0], given: 0

### Show Tags

19 Jun 2008, 08:56
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.
Intern
Joined: 05 Nov 2007
Posts: 35
Followers: 0

Kudos [?]: 2 [0], given: 0

### Show Tags

19 Jun 2008, 09:31
Hmm yah I think that works out, I don't have the proof either...@ work
Intern
Joined: 12 Mar 2008
Posts: 23
Followers: 0

Kudos [?]: 6 [0], given: 0

### Show Tags

19 Jun 2008, 10:54
How about 5 offices, 8 employees and one office can not have more than 4 employees?
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 546

Kudos [?]: 3555 [0], given: 360

### Show Tags

19 Jun 2008, 12:08
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.

$$N=4^8+C^8_1*4^7+C^8_2*4^6+C^8_3*4^5+C^8_4*4^4$$
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Joined: 12 Mar 2008
Posts: 23
Followers: 0

Kudos [?]: 6 [0], given: 0

### Show Tags

19 Jun 2008, 12:18
Can you explain especially 5C1 part...
Thanks!!
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 546

Kudos [?]: 3555 [0], given: 360

### Show Tags

19 Jun 2008, 12:27
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
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

Kudos [?]: 8 [0], given: 0

### Show Tags

19 Jun 2008, 12:30
5C1 => 5 Combination 1 => 5!/(1!(5-1)!) = 5

nCr = n!/(r!(n-r)!)
Re: PS combinatory   [#permalink] 19 Jun 2008, 12:30
Display posts from previous: Sort by