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A certain company assigns employees to offices in such a way

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

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New post 07 Nov 2016, 15:01
VeritasPrepKarishma wrote:
SoniaSaini 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


thanks in advance!!!


For each one of the 3 employees, there are two choices. He can be allotted to any one of the two offices. Hence total number ways will be 2 * 2* 2 = 8 ways



Another way to put it:

There are 3 employees in total, so the max number of people in one office is 3. And the min is 0.

Ways to distribute 3 employees (E) in 2 offices.

Office 1:

E E E
E E _
E _ _
_ _ _ (empty)

Office 2: (Same as office 1.)

E E E
E E _
E _ _
_ _ _ (empty)


Since there is no distinction between each employee for this particular question, there are 4 ways to have 3 employees fill an office. Since there are two offices, there are:
4 + 4 = 8 ways to assign 3 employees to 2 offices


Would this approach be correct? VeritasPrepKarishma
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Re: A certain company assigns employees to offices in such a way [#permalink]

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New post 19 Dec 2016, 08:58
sagarsabnis wrote:
i am still not able to understand. Can you please explain in detail?

also please tell me where i went wrong.This was my logic.

No. of people
office 1: 0|0|0|1|1|1|2|2|3
office 2: 1|2|3|0|1|2|0|1|0

this gives me 9 possible combination


Do you know how to solve this problem
In how many ways 4 distinct rings can be can be worn in 3 finger? Is it 3^4 or 4^3??
How do you decide it?

So we have some thing called reducing component here. Which ever reduces goes in the exponent. When you select one ring for any of the three fingers, the rings reduced from 4 to 3. So rings are the reducing factor. Hence no of ring = 4 will go on the exponent. There for correct answer is 3^4 and not 4^3

Similarly here Ring= Employees [Beacuase when you choose any employee to fit into any office, the office remains the same but the employee reduces. Hence employee=3 is the reducing factor}
Finger= Office

Therefore, answer will be 2^3=8
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Re: A certain company assigns employees to offices in such a way [#permalink]

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New post 21 Dec 2016, 09:09
sagarsabnis 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


We need to determine in how many ways the company can assign 3 employees to 2 different offices when some of the offices can be empty and more than one employee can be assigned to an office.

Since there are 3 people and 2 offices, we have 3 options for each office. Thus, the employees can be organized in 2^3 = 8 possible ways.

Alternative solution:

If you have trouble understanding why there should be 2^3 = 8 possible ways to assign 3 employees in 2 different offices, we can list all the possible ways one can assign 3 employees (say A, B and C) to 2 different offices (Office 1 and Office 2).

1) Office 1: A, B, C and Office 2: no one

2) Office 1: A, B and Office 2: C

3) Office 1: A, C and Office 2: B

4) Office 1: B, C and Office 2: A

5) Office 1: A and Office 2: B, C

6) Office 1: B and Office 2: A ,C

7) Office 1: C and Office 2: A, B

8) Office 1: no one and Office 2: A, B, and C

As we can see, there are 8 ways to assign 3 employees to 2 different offices.

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

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New post 27 May 2017, 04:45
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sagarsabnis 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



Each of the 3 employees has 2 choices is the best explanation.
So we have 2^3 =8 as the answer.
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Re: A certain company assigns employees to offices in such a way [#permalink]

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New post 10 Sep 2017, 04:17
I solved this in another way, though I don't know whether that's a correct one.

Considering there are 2 alternatives:

A) 3 employees to 1 office we'll have: 3C3 * 2 (since there are 2 offices).
B) 2 employees to 1 office we'll have: 3C2 * 2 (since, as above, there are 2 offices).
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Re: A certain company assigns employees to offices in such a way [#permalink]

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New post 09 Oct 2017, 11:42
VeritasPrepKarishma wrote:
ashiima wrote:
Hi,
I am kind of lost on all probability type qs :/

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


Think in this way:
There is no restriction on the offices i.e. they can be vacant, they can accommodate all 3 employees etc. But there is a restriction on the employees i.e. each one of them must get an office.

Employee 1 can get an office in 2 ways - office A or office B
Employee 2 can get an office in 2 ways - office A or office B
Employee 3 can get an office in 2 ways - office A or office B
All three can be allotted offices in 2*2*2 = 8 ways
This takes care of all cases.


Hi Karishma,
I understood the logic behind it but I always get confused when it comes to...when do we add the choices and when do we multiply the choices. Can you please explain the rationale behind it and also give a simple example?
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Re: A certain company assigns employees to offices in such a way   [#permalink] 09 Oct 2017, 11:42

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