Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be

[BTGModeratorVI]

Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

A. 240
B. 480
C. 540
D. 720
E. 840

Answer: B
Source: Kaplan

[dashingnightmare]

If there was no constraint :
Total number of ways=6*5*4*3*2*1=720

Suppose Betsy and Emily are sitting next to each other, hence we will consider them as one entity:
Total arrangements in this case=5*4*3*2*(2!)
2! is coming because betsy and emily can be arranged in 2!ways

Now, total arrangements-arrangements in which emily and betsy are sitting together will be our required answer
hence 720-240=480
Hence B.

[Brent@GMATPrepNow]

Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

A. 240
B. 480
C. 540
D. 720
E. 840

Answer: B
Source: Kaplan

Take the task of arranging the 6 students in a row and break it into stages.
NOTE: We're going to ignore the chairs and just examine how many ways there are to arrange the 6 children in a row.
The answer will be the same.

Stage 1: Arrange Arya, Chen, Daniel and Franco is a row
We can arrange n unique "objects" in a row in n! ways
So, we can arrange these 4 children in 4! ways (24 ways)
So, we can complete stage 1 in 24 ways

IMPORTANT STEP: For each of the 24 arrangements of Arya, Chen, Daniel and Franco, add a space on either side of each child.
For example, one of the possible arrangements is Chen, Daniel, Franco, Arya
So, add spaces as follows: ___Chen___Daniel___Franco___Arya___
We'll now place Betsy in a space and place Emily in a different space.
This will ENSURE that Betsy and Emily are not seated together.

Stage 2: Choose a space to place Betsy
There are 5 spaces to choose from.
So we can complete stage 2 in 5 ways

Stage 3: Choose a space to place Emily
Once we select a space in stage 2, there are 4 spaces remaining to choose from.
So we can complete stage 3 in 4 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange all 6 children) in (24)(5)(4) ways (= 480 ways)

Answer: B

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

[gentvenus]

6*5*4*3*2*1=720
5*4*3*2*(2!)=240
720-240=480

B

[Scott@TargetTestPrep]

Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

A. 240
B. 480
C. 540
D. 720
E. 840

Answer: B
Source: Kaplan

Solution:

We are given that Arya, Betsy, Chen, Daniel, Emily, and Franco are to be seated in a single row of six chairs, and that Betsy cannot sit next to Emily. We need to determine how many different arrangements of the six children are possible.

We can use the following equation:

[Total number of arrangements] = [number of arrangements with Betsy next to Emily] + [number of arrangements with Betsy not next to Emily].

Let’s rearrange the above equation as:

[Number of arrangements with Betsy not next to Emily] = [total number of arrangements] - [number of arrangements with Betsy next to Emily].

Since we are arranging 6 children, they can be arranged in 6! = 720 ways. Now we can determine how many ways to arrange them when Betsy is next to Emily. We can represent this arrangement as:

[A]-[B-E]-[C]-[D]-[F]

Notice that since Betsy is next to Emily, we consider them as 1 unit. Since there are 5 units to arrange, those units can be arranged in 5! = 120 ways. We also must account for the fact that when sitting together, Betsy and Emily can be arranged in 2! ways ([B-E] or [E-B]), so the total number of ways to arrange the children when Emily is next to Betsy is 2 x 120 = 240 ways.

Thus, the number of ways to arrange the children when Betsy is not next to Emily is 720 - 240 = 480 ways.

Answer: B