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by dtweah » Wed May 27, 2009 3:30 pm
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) 21
(C) 33
(D) 60
(E) 40
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by dumb.doofus » Wed May 27, 2009 3:53 pm
I think it should be 21...

Logic is pretty simple..

If you have one green ball, then it can be anywhere.. so 6 possible places..

If you have 2 green balls then they can be in 5 places only.. because it is mentioned that they have to be consecutive.. if you treat those two balls as one.. you'll see that they can be in 5 different positions and still be adjacent..

Same logic goes for 3, 4, 5 and 6 green balls.. we'll have 4,3,2,1 possible places respectively..

so total number of ways = 6+5+4+3+2+1 = 21..
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