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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Permutation/Combination Problem

Expert replies
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?
Join the discussion
Source: — Problem Solving |


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
Join the discussion

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
Join the discussion

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
Join the discussion