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 Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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
Start free 5-day trial
Read 559 reviewsTarget Test Prep

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

A group contains 7 boys and some girls

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members
Location: Montreal
Thanked: 1090 times
GMAT Score:780

by Ian Stewart » Mon Oct 27, 2008 3:33 pm
With 7 boys, you can choose 7C3 = 35 groups containing three boys. From n girls, you can choose nC2 = n(n-1)/2 teams containing two girls. We can thus choose (7C3)*(nC2) teams of five, containing three boys and two girls. We know that this number is equal to 525:

7C3*nC2 = 525
35*(n*(n-1))/2 = 525
n*(n-1)/2 = 15
n*(n-1) = 30
n = 6

(since n must be positive, n cannot be -5). So there are 6 girls.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 203
Joined: Wed Jun 04, 2008 1:01 am
Location: Windsor
Thanked: 5 times
GMAT Score:650

by jsl » Tue Oct 28, 2008 4:19 am
Ian Stewart wrote:From n girls, you can choose nC2 = n(n-1)/2 teams containing two girls
Thanks for the explanation Stewart! As usual, it's pragmatic and useful for the actual test!

Just had a question about the above. How did you infer that equation above? Is that something that you just knew or do you have to know some specific combinatorics equation?

I managed to get nC2 to...

n! / ( 2!(n-2)! )

but couldn't simplify further.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members
Location: Montreal
Thanked: 1090 times
GMAT Score:780

by Ian Stewart » Tue Oct 28, 2008 4:41 am
jsl wrote:
Ian Stewart wrote:
Just had a question about the above. How did you infer that equation above? Is that something that you just knew or do you have to know some specific combinatorics equation?

I managed to get nC2 to...

n! / ( 2!(n-2)! )

but couldn't simplify further.
No, I'm not using any special formula here - if you got to n! / ( 2!(n-2)! ), you were just about there. Whenever you divide one factorial by another, you'll always get a lot of cancellation. For example, (8!)/(6!) = 8*7, because

8! = 8*7*6!

so

(8!)/(6!) = (8*7*6!)/(6!) = 8*7

I was doing the same, but with n! and (n-2)!. Because

n! = n*(n-1)*(n-2)!

n!/(n-2)! = n*(n-1)*(n-2)!/(n-2)! = n*(n-1)
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 203
Joined: Wed Jun 04, 2008 1:01 am
Location: Windsor
Thanked: 5 times
GMAT Score:650

by jsl » Tue Oct 28, 2008 4:58 am
Thanks Ian! That's great - such a new and core concept for me that it's now in my notebook!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1