BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Basketball Team combos

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 58
Joined: Thu Jul 22, 2010 5:40 pm
Thanked: 1 times
by Jinglander » Wed Aug 11, 2010 10:09 am
9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?

I know how to solve this multiply the prob of coosing 6 from 3 and 2 from 3 but how about solving it with using 9! as the denominator.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Tue Aug 10, 2010 8:03 am

by fmillar » Wed Aug 11, 2010 10:51 am
Hmm... I do not see your point using 9!

because you have 6 players and 3 positions for the guards 6c3 = 20 and 3c2 for the forwards = 3

then to distribute 20 and 3 there are 60 alternatives and that should be the answer.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 58
Joined: Thu Jul 22, 2010 5:40 pm
Thanked: 1 times

by Jinglander » Wed Aug 11, 2010 11:14 am
I mean you could use the total number of teams possible ans subtract the ones that do not include 3 forward and 2 gaurds. that is what I am looking for
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1