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UGC-NET | UGC NET CS 2015 Dec – II | Question 48
  • Last Updated : 02 May, 2018

How many solutions are there for the equation x + y + z + u = 29 subject to the constraints that x ≥ 1, y ≥ 2, z ≥ 3 and u ≥ 0?
(A) 4960
(B) 2600
(C) 23751
(D) 8855


Answer: (B)

Explanation: x ≥ 1, y ≥ 2, z ≥ 3, we have to subtract these constraints from total number of choices:
i.e. 29-(1+2+3+0) = 23
(23 + 4 – 1)C23 = 2600
So, option (B) is correct.

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