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Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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27 Dec 2015, 08:46

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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?

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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27 Dec 2015, 09:57

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2

Bunuel 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

Since it is more time consuming and error prone to find ways Betsy and E not sitting together, It is better we find ways in which they will be together and then subtract from total ways..

total ways = 6!..

ways B and E will be sitting together.. take both B and E as one, then these two together with other 4 can sit in 5! ways ... Also B and E can sit within themselves in 2! ways..

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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31 Dec 2015, 06:21

1

1

Consider B and E as one entity. The arrangement of these two is possible in '2' ways

Now B&E as one entity and the other 4 children = total of 5 entities that need to be arranged = 5! ways

So, number of ways of arranging in which B and E sit together = 5! * 2

Total number of ways of arranging the 6 students = 6!

Total number of ways of arranging 6 students together - Number of ways in which B and E sit together = Total number of ways in which B and E don't sit together.

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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05 Jan 2016, 19:30

2

Hi All,

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'

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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07 Jun 2017, 10:41

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4

Bunuel 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

One more approach:

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)

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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10 Jun 2017, 07:38

2

Bunuel 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.

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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23 Oct 2017, 12:37

GMATPrepNow wrote:

Bunuel 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

One more approach:

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)

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEOS

Hi Brent,

I used another approach you teach. Seems easier.

Started with the most restrict: (1* x 4 x 4 x 3 x 2 x 1) = 96 possibilities. *means Betsy or Emily. Then, multiplied by 5 different chairs they could be seated 96 x 5 = 480.

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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23 Oct 2017, 12:53

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GC2808 wrote:

GMATPrepNow wrote:

Bunuel 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

One more approach:

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)

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEOS

Hi Brent,

I used another approach you teach. Seems easier.

Started with the most restrict: (1* x 4 x 4 x 3 x 2 x 1) = 96 possibilities. *means Betsy or Emily. Then, multiplied by 5 different chairs they could be seated 96 x 5 = 480.

Make sense? Did I miss something?

Thanks!

That seems right. However, to be sure, can you tell me what each value represents?

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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23 Oct 2017, 13:10

2

That seems right. However, to be sure, can you tell me what each value represents?

Cheers, Brent[/quote]

Let's start with Betsy. This is: 1 Then we could have any of the other: 4 (not Emily) Next: Any other 3 + Emily: 4 Next: Any other 2 + Emily: 3 Next: The other + Emily: 3 Next: Emily or the one not chosen before: 1

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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24 Oct 2017, 08:52

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GC2808 wrote:

That seems right. However, to be sure, can you tell me what each value represents?

Cheers, Brent

Let's start with Betsy. This is: 1 Then we could have any of the other: 4 (not Emily) Next: Any other 3 + Emily: 4 Next: Any other 2 + Emily: 3 Next: The other + Emily: 3 Next: Emily or the one not chosen before: 1[/quote]

Is your first step seating Betsy in one of the 5 seats, or is it selecting someone for seat #1?

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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25 Oct 2017, 13:01

GMATPrepNow wrote:

GC2808 wrote:

That seems right. However, to be sure, can you tell me what each value represents?

Cheers, Brent

Let's start with Betsy. This is: 1 Then we could have any of the other: 4 (not Emily) Next: Any other 3 + Emily: 4 Next: Any other 2 + Emily: 3 Next: The other + Emily: 3 Next: Emily or the one not chosen before: 1

Is your first step seating Betsy in one of the 5 seats, or is it selecting someone for seat #1?

Cheers, Brent[/quote]

It would be seating Betsy. That way I really know the next couldn't be Emily. When I multiply by 5 at the end it is counting all the possibilities that I could start.

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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04 Jun 2019, 16:40

I think I am finally getting it

So we have two people who cannot sit together:

There are 10 ways in which they can sit 2 people together so we multiple the probability of picking one after another by 10 -> (2/6 * 1/5) which gives us 20/30 and then we multiple this by 6! to get 480.

All the other ways listed above make no sense to me.

gmatclubot

Re: Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be
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04 Jun 2019, 16:40