JEE Main 2024 — Sets and Relations Question with Solution

Given,

$x+2y+3z=42,x,y,z\ge 0$

Now, taking cases for different value of $z$ we get,

$z=0;$ $x+2y=42$

Now, $y$ can take values $\left\{0,1,2,....21\right\}\to 22 \text{elements}$

Similarly, we get

$z=1;$ $x+2y=39\Rightarrow 20 \text{elements}$

$z=2;$ $x+2y=36\Rightarrow 19 \text{elements}$

$z=3;$ $x+2y=33\Rightarrow 17 \text{elements}$

$z=4;$ $x+2y=30\Rightarrow 16 \text{elements}$

$z=5;$ $x+2y=27\Rightarrow 14 \text{elements}$

$z=6;$ $x+2y=24\Rightarrow 13 \text{elements}$

$z=7;$ $x+2y=21\Rightarrow 11 \text{elements}$

$z=8;$ $x+2y=18\Rightarrow 10 \text{elements}$

$z=9;$ $x+2y=15\Rightarrow 8 \text{elements}$

$z=10;$ $x+2y=12\Rightarrow 7 \text{elements}$

$z=11;$ $x+2y=9\Rightarrow 5 \text{elements}$

$z=12;$ $x+2y=6\Rightarrow 4 \text{elements}$

$z=13;$ $x+2y=3\Rightarrow 2 \text{elements}$

$z=14;$ $x+2y=0\Rightarrow 1 \text{elements}$

Total : $169$