The number of integral solutions of the equation 2 x + 2 y + z = 20 where x ≥ 0 , y ≥ 0 and z ≥ 0 is 11 k then k =

2 x + 2 y = 20 − z ⇒ x + y = 10 − 2 z Coefficients of F : 0 − 2 z in ( 1 + L D + P 2 … ) 2 = 11 − 2 z Patting z = 0.2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 The number of integral solutions = 66

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The number of integral solutions of the equation $2 x + 2 y + z = 20$ where $x \geq 0, y \geq 0$ and $z \geq 0$ is $11 k$ then $k =$

Answer:

Solution:

$2 x + 2 y = 20 - z \Rightarrow x + y = 10 - \frac{z}{2}$

Coefficients of $F^{: 0 - \frac{z}{2}}$ in $(1 + L^{D} + P^{2} \dots)^{2} = 11 - \frac{z}{2}$ Patting $z = 0.2, 4, 6, 8, 10, 12, 14, 16, 18, 20$ The number of integral solutions $= 66$

Stream:JEESubject:MathematicsTopic:ProbabilitySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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