December2015 Cs Q48

0. 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?

Answer
Comment
Array

Option : 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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