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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Mar 2008 13:13:24 IST
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Algorithm for second problem:
the problem can be made simple by interpreting it in a different perspective.
Consider a, b, c as there different boxes. U have 4 red ball (red is the divisor 2 in the product), 5 green balls (green is 3) and 3 blue balls(blue is 5).
Now the problem is to place these 12 balls in 3 boxes, so that each box is filled with atleast one ball. Once again principle of inclusion and exclusion... So, the problem is same as the previous one..
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Satyaram B V,
General Secretary, Mandakini Hostel,
IIT Madras |
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Moderation Team
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