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 →

6 boxes numbered 1, 2 ...6

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members
Location: India
Thanked: 267 times
GMAT Score:760
by sanju09 » Tue Apr 05, 2011 4:27 am
There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
(A) 5
(B) 6
(C) 21
(D) 33
(E) 60
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 965
Joined: Thu Jan 28, 2010 12:52 amFollowed by: 34 members
Thanked: 156 times
GMAT Score:720

by vineeshp » Tue Apr 05, 2011 4:40 am
Case in which all 6 contain green: 1
Case in which 5 contain green: 2
Case in which 4 contain green: 3
Case in which 3 contain green: 4
Case in which 2 contain green: 5
Case in which 1 contains green: 6

21 C?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Tue Apr 05, 2011 4:41 am
sanju09 wrote:There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
(A) 5
(B) 6
(C) 21
(D) 33
(E) 60
The least no. of green balls can be 1.
If there is 1 green and 5 red balls, then there are 6 ways of placing them in 6 boxes.
If there are 2 green and reaming 4 red balls, then this can be done in 5 ways.
If there are 3 green and 3 red balls, then this can be done in 4 ways.
If there are 4 green and 2 red balls, then this can be done in 3 ways.
If there are 5 green and 1 red ball, then this can be done in 2 ways.
If there are 6 green and 0 red balls, then this can be done in 1 way.

So, total no. of ways = 6 + 5 + 4 + 3 + 2 + 1 = 21 ways

The correct answer is C.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 582
Joined: Tue Mar 08, 2011 12:48 amFollowed by: 6 members
Thanked: 61 times
GMAT Score:740

by force5 » Wed Apr 06, 2011 4:34 am
yes 21 ways.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 18
Joined: Wed Dec 22, 2010 11:31 pm

by fr743 » Wed Apr 06, 2011 6:42 am
Anurag@Gurome wrote:
sanju09 wrote:There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
(A) 5
(B) 6
(C) 21
(D) 33
(E) 60
The least no. of green balls can be 1.
If there is 1 green and 5 red balls, then there are 6 ways of placing them in 6 boxes.
If there are 2 green and reaming 4 red balls, then this can be done in 5 ways.
If there are 3 green and 3 red balls, then this can be done in 4 ways.
If there are 4 green and 2 red balls, then this can be done in 3 ways.
If there are 5 green and 1 red ball, then this can be done in 2 ways.
If there are 6 green and 0 red balls, then this can be done in 1 way.

So, total no. of ways = 6 + 5 + 4 + 3 + 2 + 1 = 21 ways

The correct answer is C.
Cant understand. How do you figure out these 6, 5 ... 1 ways?


got it. missed the word CONSECUTIVELY.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 18
Joined: Wed Mar 30, 2011 3:26 pm

by Obsessed » Wed Apr 06, 2011 9:47 am
I am lost on how to solve that one as well.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Wed Mar 15, 2017 9:59 pm

by raviksk » Wed Mar 15, 2017 10:06 pm
1st case when there is 1 green ball- {grrrrr, rgrrrr, rrgrrr, .... rrrrrg} = 6
2nd case when there are 2 green ball- {ggrrrr, rggrrr, rrggrr, rrrggr, rrrrgg } =5
3rd case when there are 3 green ball- {gggrrr, rgggrr, rrgggr, rrrggg} =4
4th case when there are 4 green ball- {ggggrr, rggggr, rrgggg} =3
5th case when there are 5 green ball- {gggggr, rggggg} =2
6th case when there are 6 green ball- {gggggg} =1

total =6+5+4+3+2+1= 21 or (6*7)/2 sum of n consecutive numbers = (n*(n+1))/2
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members
Location: East Bay all the way
Thanked: 625 times
GMAT Score:780

by Matt@VeritasPrep » Thu Mar 16, 2017 8:11 pm
Anurag@Gurome wrote:If there are 2 green and reaming 4 red balls
I'm definitely missing something here.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1