The number of non-negative integral solutions | vedclass

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Question

(B) Given the equation $x+y+z+t=10$ with constraints $x \geq 2$ and $z \geq 5$.Let $x = x' + 2$ where $x' \geq 0$.Let $z = z' + 5$ where $z' \geq 0$.S

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Given the equation $x+y+z+t=10$ with constraints $x \geq 2$ and $z \geq 5$.Let $x = x' + 2$ where $x' \geq 0$.Let $z = z' + 5$ where $z' \geq 0$.Substituting these into the equation: $(x' + 2) + y + (z' + 5) + t = 10$.$x' + y + z' + t + 7 = 10$.$x' + y + z' + t = 3$.The number of non-negative integral solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n=3$ and $r=4$.Number of solutions = $\binom{3+4-1}{4-1} = \binom{6}{3}$.$\binom{6}{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.

The number of non-negative integral solutions of the equation $x+y+z+t=10$ when $x \geq 2$ and $z \geq 5$ is:

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