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 →

Difficult Math Problem #15

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 354
Joined: Tue Jun 27, 2006 9:20 pmFollowed by: 5 members
Thanked: 11 times
by 800guy » Mon Sep 18, 2006 7:53 pm
OA coming after some people attempt an answer:

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?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Sun Sep 03, 2006 6:00 am

my suggestion

by gamemaster » Tue Sep 19, 2006 3:39 am
Total possibilites: 5!

if i arrange couple A with single B like this:

AAB

then i can put couple C in the places marked by '_' :

_A_A_B_

so ill have 4P2 = 12 options (A is man and female i assume :lol: so thats why it is 4P2)

i can make 6 like this, so i have 72 options

so total there are 72 options to arrange couples to not sit by each other out of 5!

72/120 = 3/5

i hope im right :P
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Sun Sep 17, 2006 8:08 pm

PS 15

by nguyen_thiet_thanh » Tue Sep 19, 2006 1:28 pm
let AA BB C stand for 5 people
if C sit on the first chair, 4 choices for the second chair and 2 choice for the third chair = 4*2= 8.
Same thing happens if C sit on the fifth chair = 8
Now, if C sit on the second chair, 4 choices (for the first chair) * 2 (for the third) = 8. Similarly, if C sit on the fourth chair, 4*2 = 8
finally, if C in the middle (third chair) 4 choices for the first*2 choices for the second*2 choices for the fourth = 16
Total = 8+8+8+8+16 = 48
Probability = 48/5! = 2/5 = 40%
Hope right
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 354
Joined: Tue Jun 27, 2006 9:20 pmFollowed by: 5 members
Thanked: 11 times

by 800guy » Wed Sep 20, 2006 4:57 pm
here's the OA:

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
Post new topic Post Reply
• Page 1 of 1