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There are 5 couples. If they will sit 10 chairs in a row suc

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by Max@Math Revolution » Sat May 28, 2016 7:09 pm
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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by Brent@GMATPrepNow » Sun May 29, 2016 1:01 pm
Max@Math Revolution 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
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: E
--------------------------

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
- https://www.beatthegmat.com/mouse-pellets-t274303.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/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Sun May 29, 2016 1:09 pm
Max@Math Revolution 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
Alternate approach:

Let the five couples be as follows:
AB, CD, EF, GH, IJ

Number of ways to arrange the five couples = 5! = 120.
Number of ways to arrange couple AB in their two adjacent seats = 2! = 2.
Number of ways to arrange couple CD in their two adjacent seats = 2! = 2.
Number of ways to arrange couple EF in their two adjacent seats = 2! = 2.
Number of ways to arrange couple GH in their two adjacent seats = 2! = 2.
Number of ways to arrange couple IJ in their two adjacent seats = 2! = 2.
To combine the options above, we multiply:
120*2*2*2*2*2 = 120*32 = 3840.

The correct answer is E.
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by Max@Math Revolution » Mon May 30, 2016 6:40 pm
5!(2^5)=3,840. Hence, the correct answer is E.
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