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.

permutations siblings

Expert replies
by venmic » Thu Sep 01, 2011 6:09 pm
If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

What is the logic in this question

56
Join the discussion
Source: — Problem Solving |

by GMATGuruNY » Thu Sep 01, 2011 8:08 pm
venmic wrote:If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

What is the logic in this question

56
Good = Total - Bad.

Total:
Number of combinations of 3 that can be formed from 8 people = 8C3 = 56.

Bad:
A bad committee combines one of the 4 sibling pairs with one of the 6 remaining people.
Number of options for the sibling pair = 4.
Number of options for the third committee member = 6.
To combine these options, we multiply:
Bad committees = 4*6 = 24.

Good committees = total-bad = 56-24 = 32.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by saketk » Fri Sep 02, 2011 12:40 am
venmic wrote:If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

What is the logic in this question

56
Hi --I think the answer you provided is Wrong. Read the post...

This question is similar to the one of the MGMAT question discussed on the Manhattan forum.. That question was about choosing 3 cars each of different color.

Here is the link -- https://www.manhattangmat.com/forums/the ... 12659.html

Now, Lets look at this question... First make 4 pairs

B1S1 B2S2 B3S3 B4S4

Select the first person from the group of 8 people ( 4 pair = 4*2)

This can be done in 8 WAYS, say B1 was selected.

Now, for the next selection S1 cannot participate ..(Reason -- constraint given in the question says that siblings cannot be together)

Therefore, we have only 6 people to choose from.
2nd person can be chosen in 6 WAYS.

Similarly, 3rd person can be chosen in 4 ways..

Total = 8*6*4 -- is this the answer? [WAIT]

We have to see that the ORDER of selection does not matter. So we divide the product by 3! [3! because we chose 3 people]

The correct answer to this question will be -- [spoiler](8*6*4)/3! = 32[/spoiler]
Join the discussion