Six children, Arya, Betsy, Chen, Daniel, Emily

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Fri Oct 20, 2017 10:55 pm

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

How to solve this kind of problem?

OA B

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Source: Beat The GMAT — Problem Solving | Fri Oct 20, 2017 10:55 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Oct 21, 2017 2:45 am

lheiannie07 wrote: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

Good arrangements = total possible arrangements - bad arrangements.

Total arrangements:
Number of ways to arrange the 6 children = 6! = 720.

Bad arrangements:
In a bad arrangement, B and E sit next to each other.
To count the bad arrangements, put B and E in a BLOCK, as follows:
[BE].
Now count the number of ways to arrange the 5 elements [BE] , A, C, D and F.
Number of ways to arrange the 5 elements [BE], A, C, D, and F = 5! = 120.
Since [BE] can be reversed to [EB], the result above must be doubled:
2*120 = 240.

Good arrangements:
Total possible arrangements - bad arrangements = 720-240 = 480.

The correct answer is B.

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by [email protected] » Sat Oct 21, 2017 10:48 am

Hi lheiannie07,

If you find the explanations to certain prompts to be a bit too complex, then you might be able to break a prompt down into smaller 'pieces' (and look for patterns among those pieces) and still get to the correct answer. Here, we're told that 6 children, (who I'll call A,B,C,D,E and F) are to be seated in a single row of six chairs. We're told that B cannot sit next to E. We're asked how many different arrangements of the six children are possible.

IF there were no 'restrictions', then the number of possible arrangements would be 6! = (6)(5)(4)(3)(2)(1) = 720. However, there IS a restriction (B cannot sit next to E), so we have to eliminate all of the possible arrangements that include B sitting next to E....

_ _ _ _ _ _

IF the first two spots were B and E, there would be....

B E 4 3 2 1 = 24 possible arrangements that start with BE.

In that same way, if the first two spots were E and B, there would be...

E B 4 3 2 1 = 24 possible arrangements that start with EB.

Notice how it's 24 and 24? What happens when B and E are in the second and third spots.....?

4 B E 3 2 1 = 24 possible arrangements. Using the pattern we've proven (above), there would be ANOTHER 24 arrangements when we switched the B and E...

4 E B 3 2 1 = 24 possible arrangements. This is clearly a pattern now, so we just have to count up all of the '24s'

1st/2nd spots = 24+24
2nd/3rd spots = 24+24
3rd/4th spots = 24+24
4th/5th spots = 24+24
5th/6th spots = 24+24

Total = 24(10) = 240 arrangements that would have B and E next to one another. We have to REMOVE those from the total...

720 - 240 = 480 possible arrangements that fit what we're looking for.

Final Answer: B

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

Contact Rich at [email protected]

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Mo2men: Legendary Member; Posts: 712; Joined: Fri Sep 25, 2015 4:39 amFollowed by: 5 members; Thanked: 14 times

by Mo2men » Sun Oct 22, 2017 11:06 am

Hi,

I tried to solve this question by another logic but it seems i failed at some point.

We have six places: - - - - - -

In first place: we can choose from 6 people

In second place: we have to choose from only 4 (to avoid Betsy to be next Emily)

In third place: we have to choose from only 4 (to avoid Betsy to be next Emily)

In forth place: we have only 3 to choose from.

In fifth place: we have only 2 to choose from.

In sixth place we heave only 1.

However, the above does not yield any answer. Where did I go wrong with my approach.

Thanks

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by GMATGuruNY » Mon Oct 23, 2017 3:09 am

Mo2men wrote:Hi,

I tried to solve this question by another logic but it seems i failed at some point.

We have six places: - - - - - -

In first place: we can choose from 6 people

In second place: we have to choose from only 4 (to avoid Betsy to be next Emily)

The step in red overly constrains the second slot.
If B or E is in the first slot, then neither B nor E may fill the second slot.
In this case, the step in red is valid, since only 4 of the 5 remaining children may fill the second slot.
But if A, C, D, or F is in the first slot, then any of the 5 remaining children may fill the second slot.
In this case, the step in red undercounts by 1 the number of options for the second slot.

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by Scott@TargetTestPrep » Fri Dec 15, 2017 6:56 am

lheiannie07 wrote: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

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

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Founder and CEO
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by Brent@GMATPrepNow » Fri Dec 15, 2017 7:25 am

lheiannie07 wrote: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

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 our free 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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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Wed Jan 10, 2018 12:57 am

Brent@GMATPrepNow wrote:
lheiannie07 wrote: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
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 our free 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

Thanks a lot!

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Wed Jan 10, 2018 12:58 am

Scott@TargetTestPrep wrote:
lheiannie07 wrote: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
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

Thanks a lot!

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