Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

08 Jan 2008, 07:55

3

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

53% (01:47) correct
47% (00:56) wrong based on 384 sessions

HideShow timer Statistics

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 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

Could you please show how you arrive to your answer?

We can have XXX 0 or 0 XXX XX X X XX

Essentially what that means is we have 2 possibilities where all 3 are in one room. and 6 possibilities 2(3!/2!) where 2 are in a room and one is in a room.

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

Could you please show how you arrive to your answer?

Say ABC are employees. ABC 0 0 ABC AB C BC A CA B A BC B CA C AB

8 WAYS.
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

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

Could you please show how you arrive to your answer?

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

Ans. Each employee can go into any of the two offices. Thus we have => 2 * 2 * 2 = 8

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

Show Tags

26 Feb 2012, 16:13

1

This post received KUDOS

3C3 <-- number of ways 3 employees can be chosen for office 1 + 3C2 <-- number of ways 2 employee can be chosen for office 1 + 3C1 <-- number of ways 1 employee can be chosen for office 1 + 3C0 <-- number of ways 0 employees can be chosen for office 1 = 1 + 3 + 3 + 1 = 8 D.

But, I do like walker's technique a lot better. Per walkers' solution, if there were 4 employees and 3 offices (as one of the posts points out), then the solution would be 4^3? Seem right?

3C3 <-- number of ways 3 employees can be chosen for office 1 + 3C2 <-- number of ways 2 employee can be chosen for office 1 + 3C1 <-- number of ways 1 employee can be chosen for office 1 + 3C0 <-- number of ways 0 employees can be chosen for office 1 = 1 + 3 + 3 + 1 = 8 D.

But, I do like walker's technique a lot better. Per walkers' solution, if there were 4 employees and 3 offices (as one of the posts points out), then the solution would be 4^3? Seem right?

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...