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
6 boxes numbered 1, 2 ...6
This topic has expert replies
- sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
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
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
- vineeshp
- Legendary Member
- Posts: 965
- Joined: Thu Jan 28, 2010 12:52 am
- Thanked: 156 times
- Followed by:34 members
- GMAT Score:720
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?
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.
Just telling you what I know and think. I am not the expert.
GMAT/MBA Expert
- Anurag@Gurome
- GMAT Instructor
- Posts: 3835
- Joined: Fri Apr 02, 2010 10:00 pm
- Location: Milpitas, CA
- Thanked: 1854 times
- Followed by:523 members
- GMAT Score:770
The least no. of green balls can be 1.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
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/
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/
Cant understand. How do you figure out these 6, 5 ... 1 ways?Anurag@Gurome wrote:The least no. of green balls can be 1.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
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.
got it. missed the word CONSECUTIVELY.
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
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
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
I'm definitely missing something here.Anurag@Gurome wrote:If there are 2 green and reaming 4 red balls