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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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New post Updated on: 26 Feb 2012, 17:30
5
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A
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E

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Question Stats:

58% (01:29) correct 42% (01:40) wrong based on 796 sessions

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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

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

Originally posted by JCLEONES on 08 Jan 2008, 08:55.
Last edited by Bunuel on 26 Feb 2012, 17:30, edited 1 time in total.
Topic is locked.
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Re: Combinatorics S23_6  [#permalink]

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New post 08 Jan 2008, 09:56
7
8
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]

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New post 25 Aug 2008, 10:12
6
1
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]

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New post 10 Jan 2008, 00:28
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]

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New post 25 Aug 2008, 10:26
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]

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New post 28 Aug 2008, 00:59
2
2
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]

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New post 28 Aug 2008, 01:17
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]

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New post 27 Sep 2009, 10:36
1
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]

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New post 09 Aug 2010, 17:24
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]

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New post 26 Feb 2012, 17:13
1
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]

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New post 26 Feb 2012, 17:29
1
1
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]

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New post 14 Mar 2018, 02:07
1
Responding to a pm:
Quote:
4*1*2=8

we have 4 options ( 0,1,2,or 3 employees) for the first office and 1 for the other. Multiply by 2 as we can assign employees to 2nd office first ,in which case 1st office will have only 1 option. Is this reasoning correct?



No, there is a problem in the logic used. There are 4 ways of distributing employees to the 2 offices. The order in which we distribute them in immaterial. So if we have 1 in office 1 and 2 people in office 2, it doesn't matter whether you put the 1 person in first or the 2 people in first. The final distribution is the same hence the logic of multiplying by 2 is nor correct.

Note that when you put 1 person in office 1 and 2 people in office 2, there are 3 distinct ways of doing it since the people are distinct (say A, B and C)
So A in office 1 and B, C in office 2 is different from B in office 1 and A, C in office 2.

So there is 1 way or 0 in office 1 and 3 in office 2.
3 ways of putting 1 in office 1 and 2 in office 2.
3 ways of putting 2 in office 1 and 1 in office 2.
and 1 way of putting 3 in office 1 and 0 in office 2.

That is how you get 1 + 3 + 3 + 1 = 8 cases.
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Re: A certain company assigns employees to offices in such a way &nbs [#permalink] 14 Mar 2018, 02:07
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