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UGCNETDec2015II48
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How many solutions are there for the equation $x+y+z+u=29$ subject to the constraints that $x \geq 1, \: \: y \geq 2 \: \: z \geq 3 \: \: and \: \: u \geq 0$?
4960
2600
23751
8855
ugcnetdec2015ii
discretemathematics
permutationsandcombinations
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Aug 8, 2016
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jothee
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Nov 11, 2017
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Sanjay Sharma

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2
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Best answer
x+y+z+u = 29 this is combination with repetition...
x>=1,y>=2,z>=3,u>=0 so allocate this to directly..
29(1+2+3+0) = 23
(23+41)C23 = 2600
This problem is similar to distributing 29 similar ball into 4 numbered boxes..
answered
Aug 8, 2016
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papesh
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papesh
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29  6= 23
(23 + 3) C3 = 26C3 = 2600
B is answer
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Aug 8, 2016
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Prashant.
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B
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