satishchaudharygmat wrote:

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