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permutation & comb

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permutation & comb

by romitvsingh » Wed Nov 23, 2011 10:47 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

pls explain in detail
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by shankar.ashwin » Wed Nov 23, 2011 11:02 pm
Various Possibilities are:

6 green balls - 1 way to arrange
5 green balls - 2 ways to arrange (1,2,3,4,5 or 2,3,4,5,6)
4 green balls - 3 ways to arrange (1,2,3,4 | 2,3,4,5 | 3,4,5,6)
and so on until,
1 green ball - 6 ways to arrange.

So total would be 1+2+3+4+5+6 = 21 B IMO
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by Anurag@Gurome » Wed Nov 23, 2011 11:37 pm
romitvsingh 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. 21
C. 33
D. 60
E. 6

pls explain in detail
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 B.
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