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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

by imane81 » Tue Feb 09, 2010 7:02 pm
Thank you for helping on this one

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
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by Ian Stewart » Tue Feb 09, 2010 7:09 pm
For each employee, you have 2 choices: assign him or her to Office A or assign him or her to Office B. You thus have 2*2*2 = 8 choices in total.
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by Stuart@KaplanGMAT » Tue Feb 09, 2010 7:34 pm
Ian Stewart wrote:For each employee, you have 2 choices: assign him or her to Office A or assign him or her to Office B. You thus have 2*2*2 = 8 choices in total.
The big difference you'll often see between a GMAT expert's solution and a mathematician's solution is simplicity.

Many people would jump into a question like this and immediately reach for formulae, greatly over-complicating the issue. Ian's solution is short, sweet and much much quicker than any "official" formula-based approach.

When you attack seemingly complex probability/set theory questions, always ask yourself "do I need a formula for this or can I just reason it out?" Even if your first inclination is toward a formula, you'll be amazed how often you don't need one.
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by fr743 » Tue May 24, 2011 12:18 am
2(two offices) * 2(two choices) * 2(but what does this "two" mean? could you explain that?)
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by manpsingh87 » Tue May 24, 2011 1:11 am
imane81 wrote:Thank you for helping on this one

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
another approach,, again its a logic based answer,, i hope you like it..!!

let two offices be O1 and O2 and three employee be x,y,z;

now as per the question we have;
O1 O2
x y,z
y x,z
z x,y
x,y,z 0
0 x,y,z
y,z x
x,z y
x,y z

total of 8 ways in organizing 3 persons into two offices..!!!
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by fr743 » Tue May 24, 2011 1:29 am
I've already sold it this way. I'm interested in the shown method above.
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by manpsingh87 » Tue May 24, 2011 1:44 am
fr743 wrote:I've already sold it this way. I'm interested in the shown method above.
for each person we have 2 options, either he/she will be selected for office 1, or he/she will not be selected for office 1;

hence total 2*2*2=8 ways;
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by stuffstuff » Tue May 24, 2011 10:56 am
From the answer choices it's obvious that order doesn't matter, but the question doesnt seem clear as to whether the identity of the employee matters.

If "ways that a company can assign employees" is intended to capture strictly the number of employees in the office, the answer would be different.

for example, if...
xy / z
is identical to
xz / y


Or am I the only one who thought it was a bit ambiguous?
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by djiddish98 » Tue May 24, 2011 1:10 pm
Agreed with the ambiguity. I tried attacking this with the divider method, but my answer was 4.
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by Ian Stewart » Wed May 25, 2011 2:02 am
stuffstuff wrote:From the answer choices it's obvious that order doesn't matter, but the question doesnt seem clear as to whether the identity of the employee matters.
I understand why you might find the question ambiguous, but in general, in any counting problem, you should assume that people are always distinguishable - Azeem is a different employee from Brenda, so it matters which office you assign Azeem to.

For the question to be interpreted as you did, the wording would need to make very clear that the objects being assigned to the offices were identical. You've answered a question like the following: "In how many ways could 3 identical marbles be distributed among two different boxes, if it is permitted for one box to be empty?" Here, if the first box gets one marble, we don't care which marble that is, because the marbles are identical.
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by Ovid » Sat Aug 12, 2017 7:47 am
Apologies on re-opening such an old thread. To clarify, based on my read of Ian's post, order DOES matter in this problem. What are the permutation formulas that can be used to solve this problem?

Note, I understand fundamental counting principle approach to solving the question, specifically: (2 decisions for employee #1)*(2 decisions for employee #2)*(2 decisions for employee #3)
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by Brent@GMATPrepNow » Sat Aug 12, 2017 8:10 am
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 ([spoiler]= 8 ways[/spoiler])

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Sat Aug 12, 2017 3:31 pm
Hi All,

When the answer choices to these types of questions are relatively small, you can use 'brute force' to list out all of the possibilities. Here, there are at least 4 options, but no more than 9, so if you're organized (and thorough with your work), then you should be able to list out all of the options without too much trouble.

We're asked to assign 3 employees to 2 rooms. (If we refer to the employees as A, B and C, the options would be...

0 in first room - A/B/C in the second room

1 in the first room:
A in first room - B/C in the second room
B in first room - A/C in the second room
C in first room - A/B in the second room

2 in the first room"
A/B in first room - C in the second room
A/C in first room - B in the second room
B/C in first room - A in the second room

A/B/C in first room - 0 in the second room

TOTAL OPTIONS = 1 + 3 + 3 + 1 = 8

Final Answer: D

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by Ovid » Sun Aug 13, 2017 9:49 am
Thanks, Brent and Rich.

And Brent, the links to other counting problems is helpful.

I understand the concept of applying the FCP to this question. However, I would like to clarify Ian's remarks. How can permutations, specifically, be applied to solve this question?

Reconciling FCP to permutations would be insightful.

Thanks.
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by Scott@TargetTestPrep » Fri Dec 08, 2017 12:04 pm
imane81 wrote:Thank you for helping on this one

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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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