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 →

ps22 probablity

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Legendary Member
Posts: 829
Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members
Location: INDIA
Thanked: 84 times

Re: ps22 probablity

by sudhir3127 » Wed Oct 15, 2008 4:19 am
ket_gmat wrote:hi
Its probability problem.
Please find attached problem.
Can any one suggest me good study material for probability
its D 2/5

Total arrangements = 5!= 120

both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

Thus prob.f atleast one couple together is 24+48/ 120 = 72/120= 3/5

Therefore

probability of no couple together is =1- 3/5 = 2/5

Hope its clear.. do let me know if u have any doubts,
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1159
Joined: Wed Apr 16, 2008 10:35 pm
Thanked: 56 times

by raunekk » Wed Oct 15, 2008 4:47 am
both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

hi sudhir,,, can u pls elaborate these two steps,,,,

thanks,,
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 829
Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members
Location: INDIA
Thanked: 84 times

by sudhir3127 » Wed Oct 15, 2008 6:03 am
raunekk wrote:
both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

hi sudhir,,, can u pls elaborate these two steps,,,,

thanks,,
here it goes raunekk....

assume the 2 couples are AB ,CD the single person is X

let AB and CD sit together . Assume AB and CD both as one unit
thus its (AB) and (CD) and E
number of ways is 2!*2!*3! = 24............................1

Now assume only one couple is sitting together ... for instance only AB

thus its

( AB) as one unit and CDE

4!* 2 = 48 ....( u can also write 2*(2!*2!*3!) = 48 ).....................2

hope its makes more sense now ... do let me know if u feel u need more clarification...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 871
Joined: Wed Aug 13, 2008 7:48 am
Thanked: 48 times

by stop@800 » Wed Oct 15, 2008 8:48 pm
sudhir3127 wrote:
raunekk wrote:
both couples are together = 3!*2!*2! = 24......................................1
Only one couple is together = 2*(2!*2!*3!) = 48...............................2

hi sudhir,,, can u pls elaborate these two steps,,,,

thanks,,
here it goes raunekk....

assume the 2 couples are AB ,CD the single person is X

let AB and CD sit together . Assume AB and CD both as one unit
thus its (AB) and (CD) and E
number of ways is 2!*2!*3! = 24............................1

Now assume only one couple is sitting together ... for instance only AB

thus its

( AB) as one unit and CDE

4!* 2 = 48 ....( u can also write 2*(2!*2!*3!) = 48 ).....................2
I think this will also include some cases which has already been counted in case 1
what about those?

hope its makes more sense now ... do let me know if u feel u need more clarification...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member
Thanked: 27 times
GMAT Score:660(

by 4meonly » Thu Oct 16, 2008 2:41 am
@ stop@800 agree

total - 5!

Combinations both coulples are together (AB)(E)(CD) = 3!*2*2 = 24
1st couple is together
(ABE)(CD) = 3!*2*2 = 24

2nd couple is together
(AB)(ECD) = 2*3!*2 = 24
But we alredy counted arrangement (AB)(E)(CD) in 1st statement 3!*2*2 = 24

I have some doubts as stop@800 does
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 145
Joined: Mon Sep 29, 2008 1:14 am
Thanked: 13 times

by mental » Thu Oct 16, 2008 9:50 am
Total arrangement for five people: 5! = 120

let the five people be ABCDE
AB are couple: (AB)CDE
all instances when AB sit together: 4!*2 = 48
taking AB as one unit - ie total 4 units. Also BA can arrange in 2 ways

other couple is CD: total arrangements - AB(CD)E = 4!*2 = 48

when both couples sit together = (AB)(CD)E = 3!*2*2 = 24

total arangements whenatleat one couple sits together = 48+48-24=72

because the arrangement of both sitting together is already included

ways when none of the couple sits together = 120 - 72 = 48

probability = 48/120 = 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