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
GMATBootcamp Starts Sep 21
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE BOOTCAMP

Live Online Bootcamp Class with Top GMAT Expert Chris Peckover

Sep 21 to Oct 9, 2026

Schedule
Mon to Fri · 7:00 to 10:00 PM ET
Included
Live classes + 6 months of TTP OnDemand
  • Boost your GMAT score in less than one month in a live online class
  • 6 months access to TTP OnDemand video courses included
View bootcamp & 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.

siblings

Expert replies
Source: — Problem Solving |

by krisraam » Wed Mar 25, 2009 5:58 pm
3 people have exactly 2 siblings. This means three of them are siblings.

4 people have exactly one sibling.

Total no of ways of selecting 2 people from 7 = 7C2 = 21

No of ways favorable ways.

1. If we select One from the group of 3(2 siblings) and one from the group of 4(1 sibling) = 3C1*4C1 = 12

2. If we select 2 from the group of 4 who are not siblings = 4C2( Total Selections) - 2 ( Selections with siblings) = 4

Probability = 16/21.

Thanks
raama
Join the discussion

by orel » Wed Mar 25, 2009 6:07 pm
thanks!

but i still can't understand the second step in solving a problem. can you please elaborate on that part?
Join the discussion

by krisraam » Wed Mar 25, 2009 6:55 pm
Feruza Matyakubova wrote:thanks!

but i still can't understand the second step in solving a problem. can you please elaborate on that part?
Four of them has exactly one sibling

a,b,c,d are the members of the group.

Assume that a,b and c,d are siblings.

2 people from 4 will be selected in 4C2 = 6 ways.

These 6 ways include selecting (a,b) and (c,d). We have to exclude them.

So 6 -2 = 4 ways we can select 2 people who are not sibllings.

Thanks
Raama
Join the discussion

by orel » Wed Mar 25, 2009 7:12 pm
i understand now
thank you!
Join the discussion

by Tryingmybest » Thu Mar 26, 2009 6:32 am
No of ways favorable ways.

1. If we select One from the group of 3(2 siblings) and one from the group of 4(1 sibling) = 3C1*4C1 = 12

2. If we select 2 from the group of 4 who are not siblings = 4C2( Total Selections) - 2 ( Selections with siblings) = 4


Question: In this why are we not considering 2 from group of 3 men who are not siblings??
Join the discussion

by krisraam » Thu Mar 26, 2009 6:44 am
Tryingmybest wrote:No of ways favorable ways.

1. If we select One from the group of 3(2 siblings) and one from the group of 4(1 sibling) = 3C1*4C1 = 12

2. If we select 2 from the group of 4 who are not siblings = 4C2( Total Selections) - 2 ( Selections with siblings) = 4


Question: In this why are we not considering 2 from group of 3 men who are not siblings??
Each one from that group has 2 siblings.
Like A has B and C as siblings.
B has A and C as siblings.
C has A and B as siblings.

Thanks
Raama
Join the discussion

by Tryingmybest » Thu Mar 26, 2009 7:20 am
Here is my Vision of it

A-B C- D X- Y-Z

- denotes siblings

Selecting 2 from 7 = 7 C2 = 21

Pairs which are not siblings = AC ,AD,AX,AY,AZ,BC,BD,BX,BY,BZ,CX,CY,CZ,DX,DY,DZ

So Probability = 16/21

Thanks
Join the discussion