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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Oct 2007 21:56:19 IST
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how many integral solutions of x+y+z=17 are possible if x,y,z>0?
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15 Oct 2007 18:17:59 IST
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Answer is C(19,2)
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15 Oct 2007 20:22:27 IST
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Ans) 3^(17)
= 129140163 integral solutions
raja uve have used the formula (n+r-1)C (r-1) but it wont be valid as the question specifies x,y,z > 0.
the method is assume 17 to be composed of 17 1s.
so each 1 can be put in x, y or z ie. in 3 ways.
so 17 times this process wil continue till 17 is distributed between x, y and z.
thus 3^(17).
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly, neighborhood spideyunlimited |
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15 Oct 2007 22:04:58 IST
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The answer is 16 C 2 =120.
The solution is (n-1) C (r-1) if 0 is not allowed and (n+r-1) C (r-1) if 0 is allowed.
@spideyunlimited
3^17 was a bad mistake. Whether u first put the 1 in x and then in y or
first in y and then in x , its the same. So you have counted the same cases again and again.
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15 Oct 2007 22:11:25 IST
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o ur right.. what ive done would be for dissimilar objects! that number is too large phew.. i got ur point thanks mate
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* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly, neighborhood spideyunlimited |
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