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6 boxes numbered 1, 2 ...6

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6 boxes numbered 1, 2 ...6

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
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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?
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Just telling you what I know and think. I am not the expert. :)
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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.
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by force5 » Wed Apr 06, 2011 4:34 am
yes 21 ways.
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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.
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by Obsessed » Wed Apr 06, 2011 9:47 am
I am lost on how to solve that one as well.
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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
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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.
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