Problem Solving

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

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 [spoiler]{D}[/spoiler]

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 ([spoiler]= 8 ways[/spoiler])

Cheers,
Brent

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

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

Hossain wrote:

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.

DavidG@VeritasPrep wrote:

Hossain wrote:

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.