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by confuse mind » Sat Jul 21, 2012 7:43 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. 6


Please give your complete solution
IMO - B
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by eagleeye » Sat Jul 21, 2012 9:57 pm
confuse mind 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
Please give your complete solution
IMO - B
We need to find in how many ways can we have green balls in the 6 boxes when the green balls are together. The inherent assumption here is that the balls are identical. With that in mind, let's figure this one out:

Case 1: Only one green ball. The green ball can occupy any of the 6 boxes. No. of ways = 6.

Case 2: Two green balls together (consecutively numbered boxes means that the balls are always together). Think of the 2 balls as a single entity, then we can have 5 position for the twin arrangement. (for the remaining 4 boxes, we have _ R _ R _ R _ R _, where the two green boxes can be anywhere in the 5 spaces shown among the 4 red balls in boxes. No of ways = 5.

Case 3: Three balls together, 4 spaces. No. of ways = 4.

In the same way, we get for the other 3 cases 3, 2, 1.

Hence total number of ways = 6+5+4+3+2+1 = 21.
B is correct.
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