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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] New post 08 Jan 2008, 07:55
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Question Stats:

54% (01:53) correct 46% (00:54) wrong based on 157 sessions
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

Open discussion of this question is here: a-certain-company-assigns-employees-to-offices-in-such-a-way-88936.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Feb 2012, 16:30, edited 1 time in total.
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Re: Combinatorics S23_6 [#permalink] New post 08 Jan 2008, 08:56
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D

for each employee there are two possibilities: first office and second office.
Therefore,
N=2^3=8
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Re: Combinatorics S23_6 [#permalink] New post 09 Jan 2008, 23:28
JCLEONES 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

Could you please show how you arrive to your answer?


We can have XXX 0 or 0 XXX XX X X XX

Essentially what that means is we have 2 possibilities where all 3 are in one room. and 6 possibilities 2(3!/2!) where 2 are in a room and one is in a room.

so 8
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Re: Combinatorics S23_6 [#permalink] New post 25 Aug 2008, 09:12
JCLEONES 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

Could you please show how you arrive to your answer?


Say ABC are employees.
ABC 0
0 ABC
AB C
BC A
CA B
A BC
B CA
C AB

8 WAYS.
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Re: Combinatorics S23_6 [#permalink] New post 25 Aug 2008, 09:26
JCLEONES 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

Could you please show how you arrive to your answer?


2*2*2 = 8 ways

OR

3C3 *2 + 3C1*2C2*2 = 2 + 6 = 8
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Re: Combinatorics S23_6 [#permalink] New post 27 Aug 2008, 23:59
Expert's post
x2suresh wrote:
walker wrote:
D

for each employee there are two possibilities: first office and second office.
Therefore,
N=2^3=8


Can you explain your answer?


Think about offices as baskets and employees as balls :)
For each ball there are 2 options. We have 3 balls, therefore, P=2*2*2=8
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Re: Combinatorics S23_6 [#permalink] New post 28 Aug 2008, 00:17
walker wrote:
x2suresh wrote:
walker wrote:
D

for each employee there are two possibilities: first office and second office.
Therefore,
N=2^3=8


Can you explain your answer?


Think about offices as baskets and employees as balls :)
For each ball there are 2 options. We have 3 balls, therefore, P=2*2*2=8


so if it was 3 employees and four rooms would it be simply be

4 * 4 * 4?
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Re: Combinatorics S23_6 [#permalink] New post 27 Sep 2009, 09:36
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

Ans. Each employee can go into any of the two offices. Thus we have
=> 2 * 2 * 2 = 8

D
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Re: Combinatorics S23_6 [#permalink] New post 09 Aug 2010, 16:24
mainhoon wrote:
How does this answer change if no office can be empty?


In that case it would be twice of 3C2

=> 3*2 = 6
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Re: A certain company assigns employees to offices in such a way [#permalink] New post 26 Feb 2012, 16:13
3C3 <-- number of ways 3 employees can be chosen for office 1
+ 3C2 <-- number of ways 2 employee can be chosen for office 1
+ 3C1 <-- number of ways 1 employee can be chosen for office 1
+ 3C0 <-- number of ways 0 employees can be chosen for office 1
= 1 + 3 + 3 + 1
= 8
D.

But, I do like walker's technique a lot better.
Per walkers' solution, if there were 4 employees and 3 offices (as one of the posts points out), then the solution would be 4^3? Seem right?
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Re: A certain company assigns employees to offices in such a way [#permalink] New post 26 Feb 2012, 16:29
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Expert's post
fortsill wrote:
3C3 <-- number of ways 3 employees can be chosen for office 1
+ 3C2 <-- number of ways 2 employee can be chosen for office 1
+ 3C1 <-- number of ways 1 employee can be chosen for office 1
+ 3C0 <-- number of ways 0 employees can be chosen for office 1
= 1 + 3 + 3 + 1
= 8
D.

But, I do like walker's technique a lot better.
Per walkers' solution, if there were 4 employees and 3 offices (as one of the posts points out), then the solution would be 4^3? Seem right?


Open discussion of this question is here: a-certain-company-assigns-employees-to-offices-in-such-a-way-88936.html
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Re: A certain company assigns employees to offices in such a way   [#permalink] 26 Feb 2012, 16:29
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