Mathematics · Permutations and Combinations

JEE Main 2024 — 1 February, Shift 1 — Question 22

The number of elements in the set S={(x,y,z):x,y,z∈Z,x+2y+3z=42,x,y,zS=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z ≥0\geq 0 } equals

Answer: 169

Numerical answer — enter this value.

Step-by-step solution

x+2y+3z=42,x,y,z≥0x+2 y+3 z=42, \quad x, y, z \geq 0

z=0x+2y=42⇒22\mathrm{z}=0 \quad \mathrm{x}+2 \mathrm{y}=42 \Rightarrow 22

z=1x+2y=39⇒20\mathrm{z}=1 \quad \mathrm{x}+2 \mathrm{y}=39 \Rightarrow 20

z=2x+2y=36⇒19z=2 \quad x+2 y=36 \Rightarrow 19

z=3x+2y=33⇒17z=3 \quad x+2 y=33 \Rightarrow 17

z=4x+2y=30⇒16z=4 \quad x+2 y=30 \Rightarrow 16

z=5x+2y=27⇒14z=5 \quad x+2 y=27 \Rightarrow 14

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

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

z=8x+2y=18⇒10\mathrm{z}=8 \quad \mathrm{x}+2 \mathrm{y}=18 \Rightarrow 10

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

z=10x+2y=12⇒7\mathrm{z}=10 \quad \mathrm{x}+2 \mathrm{y}=12 \Rightarrow 7

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

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

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

z=14x+2y=0⇒1\mathrm{z}=14 \quad \mathrm{x}+2 \mathrm{y}=0 \Rightarrow 1

Total : 169

Answer key and solution verified before publishing.

Practise Permutations and Combinations

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2024
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Number of integral solution of linear Equations
The number of elements in the set S=\ (x, y, z): x, y, z in Z , x+2… | JEE Main 2024 PYQ with Solution · DhiX AI