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There are 5 couples. If they will sit 10 chairs in a row such that eac

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There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 28 May 2016, 20:08
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There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?
A. 120
B. 240
C. 1,200
D. 2,460
E. 3,840

*An answer will be posted in 2 days.

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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 28 May 2016, 21:28
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MathRevolution wrote:
There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?
A. 120
B. 240
C. 1,200
D. 2,460
E. 3,840

*An answer will be posted in 2 days.


Hi,

Lets take each couple as a single identity, so total 5 places..
these 5 places can be arranged in 5! ways....
and each couple can be arranged in 2! within themselves..

so total ways = \(5!*2!*2!*2!*2!*2! = 5!*2^5 = 120*32 = 3840\)
E
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There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post Updated on: 28 May 2016, 21:34
1
MathRevolution wrote:
There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?
A. 120
B. 240
C. 1,200
D. 2,460
E. 3,840

*An answer will be posted in 2 days.


Tie the couples. Then we have 5 persons that have to be arranged in 5 places. 5! ways.

Now the couples can change position with each other. 2! ways.

5! * (2!)^2 = 120 *32 = 3840

Answer is E.

5! * 2! = 240 ways.

Originally posted by adiagr on 28 May 2016, 20:26.
Last edited by adiagr on 28 May 2016, 21:34, edited 1 time in total.
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 29 May 2016, 14:03
MathRevolution wrote:
There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?
A. 120
B. 240
C. 1,200
D. 2,460
E. 3,840

*An answer will be posted in 2 days.


Here's one approach:

Take the task of seating the 10 people and break it into stages.

Stage 1: Select someone to sit in the 1st chair
There are 10 people to choose from, so we can complete stage 1 in 10 ways

Stage 2: Select someone to sit in the 2nd chair
The person seated in the 2nd chair must be the partner of the person in the 1st chair
So we can complete this stage in 1 way.

Stage 3: Select someone to sit in the 3rd chair
There are 8 people remaining. So, we can complete stage 3 in 8 ways

Stage 4: Select someone to sit in the 4th chair
The person seated in the 4th chair must be the partner of the person in the 3rd chair
So, we can complete this stage in 1 way.

Stage 5: Select someone to sit in the 5th chair
There are 6 people remaining. So, we can complete stage 5 in 6 ways

Stage 6: Select someone to sit in the 6th chair
The person seated in the 6th chair must be the partner of the person in the 5th chair
So, we can complete this stage in 1 way.

Stage 7: Select someone to sit in the 7th chair
There are 4 people remaining. So, we can complete stage 7 in 4 ways

Stage 8: Select someone to sit in the 8th chair
The person seated in the 8th chair must be the partner of the person in the 7th chair
So, we can complete this stage in 1 way.

Stage 9: Select someone to sit in the 9th chair
There are 2 people remaining. So, we can complete stage 9 in 2 ways

Stage 10: Select someone to sit in the 10th chair
One 1 person remains.
So, we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 10 stages (and thus seat all 10 people) in (10)(1)(8)(1)(6)(1)(4)(1)(2)(1) ways ([spoiler]= 3840 ways[/spoiler])

Answer:
--------------------------

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: http://www.gmatprepnow.com/module/gmat-counting/video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Cheers,
Brent
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 30 May 2016, 19:40
5!(2^5)=3,840. Hence, the correct answer is E.
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There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 07 Jun 2016, 00:42
MathRevolution wrote:
5!(2^5)=3,840. Hence, the correct answer is E.


Could anyone perhaps explain me the logic behind the highlighted part? I went on and just multiplied 5! by 2 and still cannot understand why it would be \(2^5\)... Thank you!
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 09 Jun 2016, 09:30
fantaisie wrote:
MathRevolution wrote:
5!(2^5)=3,840. Hence, the correct answer is E.


Could anyone perhaps explain me the logic behind the highlighted part? I went on and just multiplied 5! by 2 and still cannot understand why it would be \(2^5\)... Thank you!


Each couple can be arranged in 2 ways within themselves. And there are 5 couples. So a total of 2*2*2*2*2 = 2^5 combinations.
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 09 Jun 2016, 12:32
Best way to explain this is that normally we would just arrange 5 groups/people in 5 sections as (5 * 4 * 3 * 2 * 1), grouping the couples up since they sit next two each other.
But in this case, each section/group of chairs/ couple / etc can be arranged two different ways (A1/A2 or A2/A1), so it
ends up being (10 * 8 * 6 * 4 * 2)
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 22 Jun 2016, 05:02
fantaisie wrote:
MathRevolution wrote:
5!(2^5)=3,840. Hence, the correct answer is E.


Could anyone perhaps explain me the logic behind the highlighted part? I went on and just multiplied 5! by 2 and still cannot understand why it would be \(2^5\)... Thank you!






Since the couples can exchange their seats, we mupltiply by 2. Then, since there are 5 couples, we multiply 2 five times. Hence, 2^5.
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 22 Jun 2016, 07:47
GMATPrepNow wrote:
MathRevolution wrote:
There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?
A. 120
B. 240
C. 1,200
D. 2,460
E. 3,840

*An answer will be posted in 2 days.


Here's one approach:

Take the task of seating the 10 people and break it into stages.

Stage 1: Select someone to sit in the 1st chair
There are 10 people to choose from, so we can complete stage 1 in 10 ways

Stage 2: Select someone to sit in the 2nd chair
The person seated in the 2nd chair must be the partner of the person in the 1st chair
So we can complete this stage in 1 way.

Stage 3: Select someone to sit in the 3rd chair
There are 8 people remaining. So, we can complete stage 3 in 8 ways

Stage 4: Select someone to sit in the 4th chair
The person seated in the 4th chair must be the partner of the person in the 3rd chair
So, we can complete this stage in 1 way.

Stage 5: Select someone to sit in the 5th chair
There are 6 people remaining. So, we can complete stage 5 in 6 ways

Stage 6: Select someone to sit in the 6th chair
The person seated in the 6th chair must be the partner of the person in the 5th chair
So, we can complete this stage in 1 way.

Stage 7: Select someone to sit in the 7th chair
There are 4 people remaining. So, we can complete stage 7 in 4 ways

Stage 8: Select someone to sit in the 8th chair
The person seated in the 8th chair must be the partner of the person in the 7th chair
So, we can complete this stage in 1 way.

Stage 9: Select someone to sit in the 9th chair
There are 2 people remaining. So, we can complete stage 9 in 2 ways

Stage 10: Select someone to sit in the 10th chair
One 1 person remains.
So, we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 10 stages (and thus seat all 10 people) in (10)(1)(8)(1)(6)(1)(4)(1)(2)(1) ways ([spoiler]= 3840 ways[/spoiler])

Answer:
--------------------------

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: http://www.gmatprepnow.com/module/gmat-counting/video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Cheers,
Brent



perfect !

solved this way only
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 16 May 2017, 08:07
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Bunuel: Please untag probability
MathRevolution wrote:
There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?
A. 120
B. 240
C. 1,200
D. 2,460
E. 3,840

*An answer will be posted in 2 days.

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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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New post 16 May 2017, 08:14
BrushMyQuant wrote:
Bunuel: Please untag probability
MathRevolution wrote:
There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?
A. 120
B. 240
C. 1,200
D. 2,460
E. 3,840

*An answer will be posted in 2 days.

______________
Done. Thank you.
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Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

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