Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 16 Jul 2019, 13:39

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7590
GMAT 1: 760 Q51 V42
GPA: 3.82
There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

28 May 2016, 20:08
2
10
00:00

Difficulty:

55% (hard)

Question Stats:

58% (01:21) correct 42% (01:20) wrong based on 140 sessions

### HideShow timer Statistics

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.

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 7764 Re: There are 5 couples. If they will sit 10 chairs in a row such that eac [#permalink] ### Show Tags 28 May 2016, 21:28 6 2 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 _________________ ##### General Discussion Senior Manager Joined: 18 Jan 2010 Posts: 251 There are 5 couples. If they will sit 10 chairs in a row such that eac [#permalink] ### Show Tags 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. CEO Joined: 12 Sep 2015 Posts: 3847 Location: Canada Re: There are 5 couples. If they will sit 10 chairs in a row such that eac [#permalink] ### Show Tags 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 _________________ Test confidently with gmatprepnow.com Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7590 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: There are 5 couples. If they will sit 10 chairs in a row such that eac [#permalink] ### Show Tags 30 May 2016, 19:40 5!(2^5)=3,840. Hence, the correct answer is E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Manager
Joined: 18 May 2016
Posts: 67
Concentration: Finance, International Business
GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)
There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

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!
_________________
Please kindly +Kudos if my posts or questions help you!

My debrief: Self-study: How to improve from 620(Q39,V36) to 720(Q49,V39) in 25 days!
Manager
Joined: 19 Dec 2015
Posts: 110
Location: United States
GMAT 1: 720 Q50 V38
GPA: 3.8
WE: Information Technology (Computer Software)
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

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.
Intern
Joined: 04 Feb 2016
Posts: 13
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

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)
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7590
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

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.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Senior Manager
Joined: 02 Mar 2012
Posts: 284
Schools: Schulich '16
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

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])

--------------------------

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
GMAT Tutor
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 620
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

16 May 2017, 08:07
1
Top Contributor
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.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 56244
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

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.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 11659
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac  [#permalink]

### Show Tags

18 Sep 2018, 17:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There are 5 couples. If they will sit 10 chairs in a row such that eac   [#permalink] 18 Sep 2018, 17:31
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne