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Doubt on Separator Method

This topic has 7 expert replies and 7 member replies

Doubt on Separator Method

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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. 4
B. 6
C. 7
D. 8
E. 9


I solved the question and got: 4!/3!1! ==> 4

However the answer is {D}

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theCodeToGMAT 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. 4
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)

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

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Oh, ok..

I was trying to solve the question using Separator method..

By distributing "3" employees between 2 offices..

So, one separator = (3+1)!/3! = 4.

Thanks Brent!!!

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theCodeToGMAT wrote:
Oh, ok..

I was trying to solve the question using Separator method..

By distributing "3" employees between 2 offices..

So, one separator = (3+1)!/3! = 4.

Thanks Brent!!!
The only issue with your method is that it does not treat the offices as distinct.
That is, it treats X and Y in office A and Z in office B as the same as X and Y in office B and Z in office A.
So, account for this, we just need to double your answer.

Having said that, I always begin every counting question by asking, “Can I take the required task and break it into individual stages?” If the answer is yes, I may be able to use the Fundamental Counting Principle (FCP) to solve the question.

More on this strategy here:
- http://www.beatthegmat.com/mba/2013/07/31/combinations-and-non-combinations-part-i
- http://www.beatthegmat.com/mba/2013/08/28/combinations-and-non-combinations-part-ii
- http://www.beatthegmat.com/mba/2013/09/27/does-order-matter-combinations-and-non-combinations-part-iii

Cheers,
Brent

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theCodeToGMAT 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. 4
B. 6
C. 7
D. 8
E. 9
The SEPARATOR method is great for counting the number of ways to distribute n IDENTICAL OBJECTS among r DISTINCT BOXES.
An example:
http://www.beatthegmat.com/inserting-stick-or-seperator-rule-t267423.html
Here, the objects being distributed -- the employees -- are NOT identical.
Thus, the separator method is inappropriate.

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Thanks Brent & Mitch for clarifying the doubt...

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theCodeToGMAT 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. 4
B. 6
C. 7
D. 8
E. 9


I solved the question and got: 4!/3!1! ==> 4

However the answer is {D}
Each employee can go to any of the 2 offices.
So the total number of combinations = 2*2*2 = 8

Choose D

Cheers

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Brent@GMATPrepNow wrote:
theCodeToGMAT 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. 4
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)

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775
Hi Brent,

you didn't consider the option of both the offices empty.

so answer should be 2*2*2 +1 = 9

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Hi nikhilgmat31,

The prompt states that the 3 employees have to be assigned to two different offices, so it is NOT possible that both offices would be empty.

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But prompt also says "such a way that some of the offices can be empty "

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Hi nikhilgmat31,

I agree that the wording of the prompt is 'clunky', but if there are only 2 offices, and each of the 3 employees has to be assigned to one of them, then where would they be assigned if all of the offices were empty? Logically, this doesn't make sense.

The prompt would have been clearer if it had stated "....in such a way that AN office can be empty..." Questions on the Official GMAT are almost always more clearly worded than this prompt.

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[quote="Brent@GMATPrepNow"][quote="theCodeToGMAT"]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. 4
B. 6
C. 7
D. 8
E. 9

Why can't we use counting method? I came up with answer A(4).
We can fill the offices(2) with 3 people in 4 ways as given below:

Office-1:3 and Office-2:0;
Office-1:2 and Office-2:1;
Office-1:1 and Office-2:2;
Office-0:0 and Office-2:3

Please let me know how this one can be done using counting method or we shouldn't.

Junayed Hossain

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[quote="Hossain"][quote="Brent@GMATPrepNow"]
theCodeToGMAT 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. 4
B. 6
C. 7
D. 8
E. 9

Why can't we use counting method? I came up with answer A(4).
We can fill the offices(2) with 3 people in 4 ways as given below:

Office-1:3 and Office-2:0;
Office-1:2 and Office-2:1;
Office-1:1 and Office-2:2;
Office-0:0 and Office-2:3

Please let me know how this one can be done using counting method or we shouldn't.

Junayed Hossain
You're missing a few scenarios because we have to consider which people are in which office. Imagine, for example, that there are three people: A, B, and C. Now let's take your second scenario, in which there are two people in office 1 and one person in office two.This could play out three ways

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

(Which makes sense. There are three people, so, logically, there are three different ways we can select one of them to be the lonely reject in office 2.)

The same logic would be true for the third scenario in which there is one person in office 1 and two people in office 2.

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[quote="DavidG@VeritasPrep"][quote="Hossain"]
Brent@GMATPrepNow wrote:
theCodeToGMAT 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. 4
B. 6
C. 7
D. 8
E. 9

Why can't we use counting method? I came up with answer A(4).
We can fill the offices(2) with 3 people in 4 ways as given below:

Office-1:3 and Office-2:0;
Office-1:2 and Office-2:1;
Office-1:1 and Office-2:2;
Office-0:0 and Office-2:3

Please let me know how this one can be done using counting method or we shouldn't.

Junayed Hossain
You're missing a few scenarios because we have to consider which people are in which office. Imagine, for example, that there are three people: A, B, and C. Now let's take your second scenario, in which there are two people in office 1 and one person in office two.This could play out three ways

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

(Which makes sense. There are three people, so, logically, there are three different ways we can select one of them to be the lonely reject in office 2.)

The same logic would be true for the third scenario in which there is one person in office 1 and two people in office 2.
Yes,I got it now,Thanks very much.

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This just can't be a question: there are too many ambiguities. (Can an office be empty? Are the offices distinguishable? Heck, are the employees? Very Happy)

Once we've clarified those points, the rest is formulaic, but we have to clarify those points before we can answer.

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