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

8 teams in conference

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Sun Jul 20, 2008 10:47 am
Location: new york city
Thanked: 1 times
by mberkowitz » Fri Sep 26, 2008 2:49 pm
there are 8 teams in a conference. how many games are possible if each team plays eachother twice?

OA 56

if each team played the other only once, the answer would be 8c2 or 28. in this case, bc they play eachother twice, the total number of games will be 28 times 2 or 56.

can somebody please explain the theory behind this counting method? i understand the ratoinal behind the factorial in this case, but have trouble distiguishing between cases in which team only play once, twice, and how to avoid counting a v b and b v a as seperate games...

thanks
Top
Join the discussion
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 316
Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member
Thanked: 36 times

by Morgoth » Fri Sep 26, 2008 3:00 pm
A B C D E F G H

Games played

AB AC AD AE AF AG AH----7 GAMES
BC BD BE BF BG BH--------6 GAMES
CD CE CF CG CH-----------5 GAMES
DE DF DG DH----------------4 GAMES
EF EG EH---------------------3 GAMES
FG FH-------------------------2 GAMES
GH-----------------------------1 GAME

1+2+3+4+5+6+7 = 28 GAMES

Each team plays with the other twice

28*2 = 56 Games

Hope its clear.
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Sun Jul 20, 2008 10:47 am
Location: new york city
Thanked: 1 times

by mberkowitz » Fri Sep 26, 2008 3:05 pm
appreciate that, i guess the best method to learn these ones is to picture it like that, i just have trouble going from there to the factorial. i.e. thats a great method if you have 4 mins, but clearly not the fastest option.
Top
Join the discussion
User avatar
Master | Next Rank: 500 Posts
Posts: 316
Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member
Thanked: 36 times

by Morgoth » Fri Sep 26, 2008 3:17 pm
mberkowitz wrote:appreciate that, i guess the best method to learn these ones is to picture it like that, i just have trouble going from there to the factorial. i.e. thats a great method if you have 4 mins, but clearly not the fastest option.
I thought you were looking for clarity rather than which method to follow.

If you are looking for the method its hard to beat 8C2, which is the most effective on such type of questions.
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Sun Jul 20, 2008 10:47 am
Location: new york city
Thanked: 1 times

by mberkowitz » Fri Sep 26, 2008 4:05 pm
i was looking for an explanation of the method in this problem. in other words this problem stumped me even though i know the method and can usually execute it well, as i find the method arbitrary at times.

thanks for the help.
Top
Join the discussion
User avatar
Legendary Member
Posts: 871
Joined: Wed Aug 13, 2008 7:48 am
Thanked: 48 times

by stop@800 » Sat Sep 27, 2008 3:24 am
8c2 is clear
when a person is playing once with everyother

so 2 * 8c2 means each person played twice.

Perhaps I could not understand your problem well.....
please help me understanding your doubt.

Thanks
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Sun Jul 20, 2008 10:47 am
Location: new york city
Thanked: 1 times

by mberkowitz » Sat Sep 27, 2008 4:31 am
i was having trouble getting 8c2 quickly bc i was tending to worry about counting teams more than once (counting a v b and b v a as two seperate games). id imagine learning that property as is is the best way for me to proceed. thanks.
Top
Join the discussion