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A certain company assigns employees to offices in such a way

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New post 19 Dec 2016, 09:58
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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New post 21 Dec 2016, 10:09
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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New post 27 May 2017, 05:45
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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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New post 10 Sep 2017, 05:17
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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New post 09 Oct 2017, 12:42
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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New post 26 Feb 2018, 04:18
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

IMO is D as we know each employ has 2 choices between offices then e1=2 e2=2 and e3=2 then 2^3=8
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New post 13 Mar 2018, 10:51
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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


Let X, Y and Z be the 3 employees.
Let A and B be the 2 offices.


Take the task of assigning the employees and break it into stages.

Stage 1: Assign employee X to an office
There two options (office A or office B), so we can complete stage 1 in 2 ways

Stage 2: Assign employee Y to an office
There two options (office A or office B), so we can complete stage 2 in 2 ways

Stage 3: Assign employee Z to an office
There two options (office A or office B), so we can complete stage 3 in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus assign all employees to offices) in (2)(2)(2) ways (= 8 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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New post 01 May 2018, 06:31
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


pushpitkc

here is my solution to this question:

\(3C1 *2C2 = 6*1\) ( # of ways to choose 1 employee from 3 and # of ways to choose 2 employees from 2)

\(3C2 *1C1 = 3*1\) ( # of ways to choose 2 employees from 3 and # of ways to choose 1 employees from 1)

\(6+3 = 9\) :?

if this is purely combinatorics questions why doesnt our approach work :) please explain :)
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New post 01 May 2018, 11:24
1
Hi dave13

The major mistake that you have made in this problem is the question stem reads
"One of the offices can have zero employees" also - meaning the other office has full capacity.

There are 2 possibilities when the 3 employees are in either rooms.

Now coming to the mistake in your solution
\(3C1 *2C2 = 6*1\) ( # of ways to choose 1 employee from 3 and # of ways to choose 2 employees from 2)

Here, \(3C1*2C2 = 3*1 = 3\). This makes the total possibilities 3+3 = 6

Therefore, the number of possibilities including the case that you missed is 2+6 = 8(Option D)

Hope this helps you!
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New post 03 May 2018, 08:48
pushpitkc wrote:
Hi dave13

The major mistake that you have made in this problem is the question stem reads
"One of the offices can have zero employees" also - meaning the other office has full capacity.

There are 2 possibilities when the 3 employees are in either rooms.

Now coming to the mistake in your solution
\(3C1 *2C2 = 6*1\) ( # of ways to choose 1 employee from 3 and # of ways to choose 2 employees from 2)

Here, \(3C1*2C2 = 3*1 = 3\). This makes the total possibilities 3+3 = 6

Therefore, the number of possibilities including the case that you missed is 2+6 = 8(Option D)

Hope this helps you!



Hi pushpitkc , many thanks :)

can you please confirm my solution below is it correct now ? :)

\(3C1 *2C2 = 3*1 = 3\) ( # of ways to choose 1 employee from 3 and # of ways to choose 2 employees from 2)

\(3C2 *1C1 = 3*1 =3\) ( # of ways to choose 2 employees from 3 and # of ways to choose 1 employees from 1)

\(3C3 *0C0 = 3*0 = 0\) ( # of ways to choose 3 employees from 3 and # of ways to choose 0 employees from 0) (since the question stem reads
"One of the offices can have zero employees")

Total number of ways: \(3+3+0= 6\)
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New post 03 May 2018, 10:39
1
dave13 wrote:
pushpitkc wrote:
Hi dave13

The major mistake that you have made in this problem is the question stem reads
"One of the offices can have zero employees" also - meaning the other office has full capacity.

There are 2 possibilities when the 3 employees are in either rooms.

Now coming to the mistake in your solution
\(3C1 *2C2 = 6*1\) ( # of ways to choose 1 employee from 3 and # of ways to choose 2 employees from 2)

Here, \(3C1*2C2 = 3*1 = 3\). This makes the total possibilities 3+3 = 6

Therefore, the number of possibilities including the case that you missed is 2+6 = 8(Option D)

Hope this helps you!


Hi pushpitkc , many thanks :)

can you please confirm my solution below is it correct now ? :)

\(3C1 *2C2 = 3*1 = 3\) ( # of ways to choose 1 employee from 3 and # of ways to choose 2 employees from 2)

\(3C2 *1C1 = 3*1 =3\) ( # of ways to choose 2 employees from 3 and # of ways to choose 1 employees from 1)

\(3C3 *0C0 = 3*0 = 0\) ( # of ways to choose 3 employees from 3 and # of ways to choose 0 employees from 0) (since the question stem reads
"One of the offices can have zero employees")

Total number of ways: \(3+3+0= 6\)



Hi dave13

I think this is how this problem can be solved using this method

First room - 2 employees, Second room - 1 employee: 3c2*1c1 = 3*1 = 3
First room - 1 employee, Second room - 2 employee: 3c1*2c2 = 3*1 = 3
3 can be placed in either room - 3c3*2 = 1*2 = 2

Total possibilities: 3+3+2 = 8

Alternative method(already suggested)

All three employees have 2 options of being placed in an office - Office 1 or 2.
Therefore, we will have (2)(2)(2) or 8 ways in which the employees can be placed in the office.

Hope this helps you!
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Re: A certain company assigns employees to offices in such a way  [#permalink]

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New post 13 May 2018, 22:51
Bunuel wrote:
sagarsabnis wrote:
A certain company assigns employees to offices in such a way thatsome 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 three employee can be assigned to either of the two offices, meaning that each employee has 2 choices --> 2*2*2=2^3=8.

Answer: D.


Hi Bunuel,
thanks for the explanation.
Initially I got confused from the above highlighted line.
And I did 3*3*3 because a person can either select 1 or 2nd or none.
Kindly please help me with this.

Thanks
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New post 21 Sep 2018, 17:05
Hi ,
well this problem has provided some information initially to and the question is actually in last part.

HOW CAN YOU ASSIGN 3 people to 2 offices" this is similar like how can you assign 3 rings to 2 fingers or similar on the lines of how can you post 7 letters to 5 different post boxes.

Say we have two spaces ___I____ and ___II_______ and 3 employees (A, B C)

A has two options either space 1 or space 2 So 2 ways
B has two options either space 1 or space 2 So 2 ways
C has two options either space 1 or space 2 So 2 ways

Altogether a has two and b has two and c has two = 2*2*2= 8 Ways.

Now try the letterbox problem and see if you got the concept .

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