Hi Roland2rule,

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

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorRO 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

OA is D
Can an expert give a mathematical approach to, solve this question? Thanks in anticipation

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 2 options for each employee. Thus, the employees can be organized in 2 x 2 x 2 = 2^3 = 8 possible ways.

Answer: D

BTGmoderatorRO 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

OA is D
Can an expert give a mathematical approach to, solve this question? Thanks in anticipation

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. For more information about the FCP, watch this 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