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
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
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.

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.

Counting question (Source GMAT PREP)

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Mon Sep 10, 2007 5:18 am
by mach81 » Wed Sep 12, 2007 5:02 pm
Would appreciate if somebody could help me with these two COUNTING questions (I guess the factorial formaula can be used, but HOW?)

1) There are 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (Ans = 100)

2)All the stocks on the OTC are designated by either 4 letters OR a 5 letters code, using 26 letters of the alphabet. What are the maximum number of stocks that can be represented (And = 27((26)^4)

Thanks
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 71
Joined: Wed Aug 22, 2007 12:29 pmFollowed by: 1 member
Location: USA
Thanked: 12 times

Answer to question 1

by wizardofwashington » Wed Sep 12, 2007 8:55 pm
Let us find how many 3-member teams can be formed from a total of 10 people (4+6). => 10 C3 = 120

Now, with the question statement "at least one senior partner must be present in these teams", we should calculate the # of possibilities without any senior partner (only juniors) and subtract this number from the 120 we calculated above.

# of possibilities without any senior partner = # of possibilities with only junior partners

Therefore, # of combinations of teams fully populated with Junior members - 6C3 = 20

So, the answer is 120-20 = 100.
A falling tree resounds... but a forest grows in silence...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 71
Joined: Wed Aug 22, 2007 12:29 pmFollowed by: 1 member
Location: USA
Thanked: 12 times

Re: Counting question (Source GMAT PREP)

by wizardofwashington » Wed Sep 12, 2007 9:11 pm
mach81 wrote:
2)All the stocks on the OTC are designated by either 4 letters OR a 5 letters code, using 26 letters of the alphabet. What are the maximum number of stocks that can be represented (And = 27((26)^4)
Question 2 has already been discussed on this forum. Please check the following link..

https://www.beatthegmat.com/viewtopic.php?p=19992#19992
A falling tree resounds... but a forest grows in silence...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 321
Joined: Tue Aug 28, 2007 5:42 am
Thanked: 1 times

by kajcha » Thu Sep 13, 2007 5:46 am
My approach for first question

Possible combinations SJJ, SSJ, SSS

4C1*6C2+4C2*6C1+4C3 = 100
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1