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When \mathrm{x=1, y=2, z=3}, how many integral solutions are there for \mathrm{x+y+z+w=29} where \mathrm{x \geq 1, y \geq 2, z \geq 3} and \mathrm{w \geq 0}?

Option: 1

2290


Option: 2

1500


Option: 3

2600


Option: 4

3000


Answers (1)

best_answer

\mathrm{x+y+z+w=29}\:- \text{Equation (1)}
\mathrm{x \geq 1, y \geq 2, z \geq 3} and \mathrm{w \geq 0}

Which can also be written as \mathrm{x-1 \geq 0, y-2 \geq 0, z-3 \geq 0} and \mathrm{w \geq 0}

\mathrm{x_{1}=x-1, x_{2}=y-2, x_{3}=z-3}
\mathrm{x=x_{1}+1, y=x_{2}+2, z=x_{3}+3}
\mathrm{x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0, w \geq 0}

Using it in equation (1)
\mathrm{x_{1}+x_{2}+x_{3}+w=23}

Then the number of solutions for this will be \mathrm{{ }^{23+4-1} C_{4-1}={ }^{26} C_{3}}
\mathrm{=2600}

Option (c) is correct.

Posted by

Pankaj

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