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QJEEM24Ma2215: JEE Main 2024 Mathematics - Class 11 JEEM24Ma2215 Solution: The number of elements in the set equals __________.

The number of elements in the set S={(x,y,z):x,y,z∈Z,x+2y+3z=42,x,y,z⩾0}\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\} equals __________.

Step-by-Step Verified Solution

x+2y+3z=42x+2 y+3 z=42

x,y,z≥0x, y, z \geq 0

as

z=0amp;x+2y=42⇒22 cases z=1amp;x+2y=39⇒20 cases z=2amp;x+2y=36⇒19 cases z=3amp;x+2y=33⇒17 cases z=4amp;x+2y=30⇒16 cases z=5amp;x+2y=27⇒14 cases \begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\\\ z=5 & x+2 y=27 \Rightarrow 14 \text { cases }\end{array}

z=6x+2y=24⇒13z=6 \quad x+2 y=24 \Rightarrow 13 cases

z=7x+2y=21⇒11z=7 \quad x+2 y=21 \Rightarrow 11 cases

z=8x+2y=18⇒10z=8 \quad x+2 y=18 \Rightarrow 10 cases

z=9x+2y=15⇒8z=9 \quad x+2 y=15 \Rightarrow 8 cases

z=10x+2y=12⇒7z=10 x+2 y=12 \Rightarrow 7 cases

z=11x+2y=9⇒5z=11 x+2 y=9 \Rightarrow 5 cases

z=12x+2y=6⇒4z=12 x+2 y=6 \Rightarrow 4 cases

z=13x+2y=3⇒2z=13 x+2 y=3 \Rightarrow 2 cases

z=14x+2y=0⇒1z=14 x+2 y=0 \Rightarrow 1 case

Total = 169

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