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
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
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.

130-point 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.

Combinatorics #6

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Community Manager
Posts: 1537
Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members
Thanked: 653 times
by papgust » Sun Oct 18, 2009 5:18 am
Several teams take part in a competition, each of which must play one game with all the other teams. How many teams took part in the competition if they played 45 games in all?

A. 5
B. 10
C. 15
D. 20

Solution and explanation pls?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Jun 18, 2009 5:33 am
Thanked: 35 times

by NikolayZ » Sun Oct 18, 2009 8:00 am
Hi there !
is the answer is B?

I did it without combinatorics. Try solving backwards from the answers.
So, if there were 5 teams.

1st must play with - 2 3 4 5
2nd -3 4 5
3d -4 5
4th - 5
5 (already played with all possible teams)
clearly - there are less games played than 45.

Try 10 the same way. You ll get 45 games total.
9+8+7+6+5+4+3+2+1=45
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 1537
Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members
Thanked: 653 times

by papgust » Sun Oct 18, 2009 8:13 am
Hi NikolayZ,

Thanks for the solution. The answer is indeed 10. But i'm looking to solve this problem by combinatorics theory. Even i solved this by backsolving. I'm looking to build my fundamentals in combinatorics as i'm very weak in this subject.

Can you solve this prob using combinations and help me understand how the theory is applied to this kind of a prob?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1