Sets & RelationseasyPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Non-negative Solutions of a+b+2c=22: n(A) = 144 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let A={(a,b,c):a,b,c are non-negative integers and a+b+2c=22}A=\{(a,b,c):a,b,c\ \text{are non-negative integers and}\ a+b+2c=22\}. Then n(A)n(A) is equal to
A121121
B124124
C144144correct
D169169
Solution
Step 1: Fix cc; then a+b=22−2ca+b=22-2c, which has (22−2c)+1=23−2c(22-2c)+1=23-2c non-negative integer solutions. Step 2: cc ranges from 00 to 1111 (so that 22−2c≥022-2c\ge0):
c=0: 23,c=1: 21,c=2: 19, …, c=11: 1.c=0:\ 23,\quad c=1:\ 21,\quad c=2:\ 19,\ \ldots,\ c=11:\ 1.
 Step 3: Sum the odd numbers 23+21+19+⋯+123+21+19+\cdots+1 (12 terms):
1+3+5+⋯+23=122=144.1+3+5+\cdots+23=12^2=144.
 Correct answer: (3)
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