Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart 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
Start free 5-day trial
Read 563 reviewsTarget Test Prep

2 three-person subgroups

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members
Location: India
Thanked: 267 times
GMAT Score:760
by sanju09 » Wed Mar 02, 2011 5:05 am
Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
The mind is everything. What you think you become. –Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Join the discussion
Source: — Problem Solving |
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members
Thanked: 127 times

by Night reader » Wed Mar 02, 2011 5:38 am
3(Bob)*2(Allen)*(some one else) OUT OF 6C3 --> 6/20=3/10=0.3 (30%)

IOM B
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
Top
Join the discussion
User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members
Thanked: 72 times

by manpsingh87 » Wed Mar 02, 2011 9:45 am
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
IMO A..!!!
Total no. of ways of forming sub groups; 6!/3!*(2!)^3 = 15

Total no. of favorable cases; 3

hence required probability or percentage is 3/15 = 1/5 = 20% [spoiler]hence A..!!![/spoiler]

Please post correct answer...!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
Top
Join the discussion
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Wed Mar 02, 2011 10:03 am
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
Number of 3-member groups that include Bob:
Bob can be combined with 5C2 = 10 other pairs.

Number of 3-member groups that include both Bob and Allen:
Bob and Allen can be combined with 4C1 = 4 other people.

Out of all the groups with Bob, 4/10 = 40% will also include Allen.

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
Top
Join the discussion
GMAT Instructor
Posts: 613
Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members
Location: madrid
Thanked: 171 times
GMAT Score:790

by kevincanspain » Wed Mar 02, 2011 2:33 pm
sanju09 wrote:Allen and Bob are among the six members to participate in a debate. If the group is to be split up into 2 three-person subgroups, what percent of all the possible subgroups that include Bob also include Allen?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


[spoiler]made up[/spoiler]
In other words, what is the probability that Allen will be one of the two people who accompany Bob? Since apart from Bob, there are 5 people, the required probability is 2/5
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
Top
Join the discussion
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Fri Dec 11, 2009 10:15 am
Thanked: 2 times

by anu009 » Sat Mar 05, 2011 8:58 pm
Hello Instructors

The probabilty & combinatin questions are really confusing for me.Each time there is a different way to solve the problem to get to the correct solution.

I was trying to do this with basic concepts of Single source order doesnt matter n!/(n-r)!*r! which was giving 6C3 = 15 and the favourable cases = 3. SO favourable/total probabilyt = 3/15 * 100 = 20% but the answer is different.

When i saw your solution i found it rt but i am not sure where was i wrong.Can you help me with this and so that the general rule of solving the questions works everywhere.

Thank you.
Top
Join the discussion

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Sat Mar 05, 2011 9:22 pm
anu009 wrote:Hello Instructors

The probabilty & combinatin questions are really confusing for me.Each time there is a different way to solve the problem to get to the correct solution.

I was trying to do this with basic concepts of Single source order doesnt matter n!/(n-r)!*r! which was giving 6C3 = 15 and the favourable cases = 3. SO favourable/total probabilyt = 3/15 * 100 = 20% but the answer is different.

When i saw your solution i found it rt but i am not sure where was i wrong.Can you help me with this and so that the general rule of solving the questions works everywhere.

Thank you.
Probability is (no. of favorable cases)/(total no. of all possible cases).

Here, what you have taken as "total no. of all possible cases" is "total ways of selecting a group of 3 from among 6 people".
Read the question again.
It is "what percent of all possible subgroups that include Bob".
So, we need to estimate as "total no. of all possible cases", the number of groups of 3, that always have Bob.
This can be done in 5C2 or 10 ways. ( this is because leaving Bob, we have to select other 2 from remaining 5).
Also "no. of favorable cases" is number of ways we can select groups of 3 which have Bob and Allen both.
This can be done in 4C1 or 4 ways (this is because we have to select 1 more person from remaining 4 apart from Bob and Allen).
So required prob. is 4/10 = 40%.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Fri Dec 11, 2009 10:15 am
Thanked: 2 times

by anu009 » Sat Mar 05, 2011 9:31 pm
Got it. Thank You so much.
Top
Join the discussion
Legendary Member
Posts: 965
Joined: Thu Jan 28, 2010 12:52 amFollowed by: 34 members
Thanked: 156 times
GMAT Score:720

by vineeshp » Sat Mar 05, 2011 9:45 pm
kevincanspain wrote:
In other words, what is the probability that Allen will be one of the two people who accompany Bob? Since apart from Bob, there are 5 people, the required probability is 2/5
Kevin,
Can you explain your solution a bit more?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
Top
Join the discussion