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Red and Green boxes

Expert replies
by harsh.champ » Tue Feb 09, 2010 4:00 am
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)55

The OA is B.
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Source: — Problem Solving |

by shashank.ism » Tue Feb 09, 2010 4:59 am
harsh.champ 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)55

The OA is B.
GRRRRR, RGRRRR, RRGRRR, RRRGRR, RRRRGR, RRRRRG
GGRRRR, RGGRRR, RRGGRR, RRRGGR, RRRRGG
GGGRRR, RGGGRR, RRGGGR, RRRGGG
GGGGRR, RGGGGR, RRGGGG
GGGGGR, RGGGGG
GGGGGG
Hence 21 ways.

Ans is B
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by ajith » Tue Feb 09, 2010 6:03 am
harsh.champ 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)55

The OA is B.
if first is green there are 6 options - all six G, 5G and green....
If first is red and second is green there are 5 options
....
so on
Total no of options = 6+5+4.... +1 = 7*6/2 = 21
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