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satishchaudharygmat
There are 6 boxes numbered 1, 2 ,3, 4, 5, 6 Each box is to be filled up either with red or green ball in a such way that at lest 1 box contains 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 21
C 33
D 60
E 40

Hi!
The topic should contain the first few words of the question. Requesting the author and the moderators to fix the issue.

As far as the question is concerned:
If there is one box that contains a green ball then there will be 6 ways to do it: 1, 2, 3, 4, 5, or 6
If there are two boxes that contain green ball(s) then there will be 5 ways to do it: (1,2) (2,3) (3,4) (4,5) or (5,6)
If there are three boxes that contain green ball(s) then there will be 4 ways to do it: (1,2,3) (2,3,4) (3,4,5) or (4,5,6)
If there are four boxes that contain green ball(s) then there will be 3 ways to do it: (1,2,3,4) (2,3,4,5) or (3,4,5,6)
If there are five boxes that contain green ball(s) then there will be 2 ways to do it: (1,2,3,4,5) or (2,3,4,5,6)
If there are six boxes that contain green ball(s) then there will be 1 way to do it: (1,2,3,4,5,6)

So in total 1+2+3+4+5+6 = 21 ways
Hence, answer B
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satishchaudharygmat
There are 6 boxes numbered 1, 2 ,3, 4, 5, 6 Each box is to be filled up either with red or green ball in a such way that at lest 1 box contains 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 21
C 33
D 60
E 40

Ways to put green ball in only one box= 6
Ways to put green ball in 2 consecutively numbered boxes= 5
Ways to put green ball in 3 consecutively numbered boxes= 4
Ways to put green ball in 4 consecutively numbered boxes= 3
Ways to put green ball in 5 consecutively numbered boxes= 2
Ways to put green ball in 6 consecutively numbered boxes= 1

Total ways= 21

B is the answer
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Various ways are as under.

GGGGGG - 1 way

GGGGGR: Bundle all 5 G's together. This bundle and R have to be placed. Can only be done in 2 ways

GGGGRR: Bundle all 4 G's together. This bundle and 2 R's have to be distributed. Total 3 articles are there. This can be done in (3!/2!= 3 ways)

[There is a bundle of G's and 2 R's are similar. That is why we are dividing by 2!]

GGGRRR: Bundle all 3 G's together. This bundle and 3 R's have to be distributed. Total 4 articles are there. This can be done in (4!/3!= 4 ways)

[There is a bundle of G's and 3 R's are similar. That is why we are dividing by 3!]

GGRRRR: Bundle 2 G's together. This bundle and 4 R's have to be distributed. Total 5 articles are there. This can be done in (5!/4!= 5 ways)

[There is a bundle of G's and 4 R's are similar. That is why we are dividing by 4!]

GRRRRR: Now there is only one G. This can be kept anywhere. 6 articles are there, 5 of one type 1 of a different type. These can be arranged in (6!/5!= 6 ways)

so 1+2+3+4+5+6 = 21 ways


Option B is correct.
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