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 →

Permutation/Combination Problem

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Tue Jun 05, 2007 1:29 pm
by dzelkas » Sun May 25, 2008 11:47 am
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?


a) 8
b) 24
c) 32
d) 56
e) 80


correct answer is C. Can someone please help explain the solution?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 192
Joined: Tue Apr 29, 2008 7:41 am
Thanked: 14 times

by aatech » Sun May 25, 2008 7:58 pm
https://www.beatthegmat.com/committee-num-t11273.html
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 1223
Joined: Thu May 01, 2008 3:29 pmFollowed by: 15 members
Location: Los Angeles, CA
Thanked: 185 times

by VP_Jim » Sun May 25, 2008 9:08 pm
I was thinking that this question sounds familiar... yes, it's pretty much the exact same set up as the question aatech linked to. My explanation for that one still stands here: just think about it logically. In my opinion, equations tend to get students in trouble.
Jim S. | GMAT Instructor | Veritas Prep
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 24
Joined: Mon Mar 31, 2008 4:11 am

by mandy12 » Sun May 25, 2008 11:55 pm
One approach to solve this question is -

total no of ways of selecting 3 people from 8 = 8C3 = 56

Now if the siblings are present in the combination then 2 places are booked and only one place is left to be filled....

1*1*6 for each sibling pair..We have 4 pairs ...so total number of combinations in which sibling pair will be present is -

4*1*1*6 = 24

Hence the number of combinations in which the sibling pair will not be presnet = 56 - 24 = 32
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 42
Joined: Fri Mar 14, 2008 1:56 am
Location: India
Thanked: 4 times
GMAT Score:730

by getneonow » Mon May 26, 2008 12:37 am
2* 4C3 ( 3 boys or gals committe) + 2 * 4C2 * 2C1 ( 2 boys and one gal or 2 gals and one boy) = 32


Neo
MBA : My passion and My pursuit
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1