BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

simple probability question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members
Location: India
Thanked: 64 times
GMAT Score:760

by harsh.champ » Fri Feb 19, 2010 3:43 am
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Let the couples be A1,A2 and B1,B2.

Now,let one A ,one B and the single person be S be seated - 3! x 2C1 x 2C1.

So,the other two A and B can be seated in 2 x 2 ways.

Hence, the answer would be 96 ways.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members
Thanked: 137 times

by thephoenix » Fri Feb 19, 2010 3:46 am
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit)
Ways for second couple = 2*4!
These cases include an extra case of both couples sitting together
Ways in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2
units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3!
Ways)
Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4!
Total ways to arrange the 5 ppl = 5!
Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5
Thus prob of none seated together = 1 - 3/5 = 2/5
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members
Thanked: 137 times

by thephoenix » Fri Feb 19, 2010 3:48 am
harsh.champ wrote:
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Let the couples be A1,A2 and B1,B2.

Now,let one A ,one B and the single person be S be seated - 3! x 2C1 x 2C1.

So,the other two A and B can be seated in 2 x 2 ways.

Hence, the answer would be 96 ways.
the q is abt probabilty
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members
Location: India
Thanked: 64 times
GMAT Score:760

by harsh.champ » Fri Feb 19, 2010 4:47 am
thephoenix wrote:
harsh.champ wrote:
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Let the couples be A1,A2 and B1,B2.

Now,let one A ,one B and the single person be S be seated - 3! x 2C1 x 2C1.

So,the other two A and B can be seated in 2 x 2 ways.

Hence, the answer would be 96 ways.
the q is abt probabilty
OIC.
I forgot to consider the sample space.
[If the option choices had been given,I could have rectified my mistake then and there only-This is also one of the imp. reasons why option choices should be given alongwith the question]
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1