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.

combination problem

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 74
Joined: Tue Nov 09, 2010 11:33 pm
by sukh » Mon Aug 08, 2011 6:15 am
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes. (A) 24 (B) 12 (C) 7 (D) 6 (E) 5
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Quantitative Reasoning |
User avatar
Legendary Member
Posts: 1255
Joined: Fri Nov 07, 2008 2:08 pmFollowed by: 90 members
Location: St. Louis
Thanked: 312 times

by Tani » Mon Aug 08, 2011 9:33 am
Since order doesn't matter you will use combinations. The simplest way to allow for a single color is to think of a single color designation as a color plus no color. Then you can consider "no color" as an additional color.

You can then analyze the number of possibilities with 5 colors by using the combination formula for 6 (5 colors plus no color)

6!/(2!*4!) = 15. Enough - the answer is 5 colors.

Looked at another way, combinations of two colors starting with 5 will give you 5!/(2! * 3!) = 10. That's ten combinations of two colors. You can also use any one of the five colors by itself - giving you again 15 possibilities.
Tani Wolff
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 74
Joined: Tue Nov 09, 2010 11:33 pm

by sukh » Tue Aug 09, 2011 5:05 am
answer is 5 , i didnt get it - calculated 15 , how does it get reduced to 5
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1255
Joined: Fri Nov 07, 2008 2:08 pmFollowed by: 90 members
Location: St. Louis
Thanked: 312 times

by Tani » Tue Aug 09, 2011 7:55 am
5 colors will give you 15 alternatives. That is the smallest number you need to have at least 12 combinations available.
Tani Wolff
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1