Mathematics · Permutations and Combinations

JEE Main 2026 — 4 April, Evening Shift — Question 41

Let $A = \{(a, b, c) : a, b, c$ are non-negative integers and $a + b + 2c = 22\}$ . Then $n(A)$ is equal to:

Option A:
$121$
Option B:
$124$
Option C:
$144$
Correct
Option D:
$169$

Answer: C

Step-by-step solution

$\mathrm{c}=0 \Rightarrow \mathrm{a}+\mathrm{b}=22 \rightarrow{ }^{22+2-1} \mathrm{C}_{2-1}={ }^{23} \mathrm{C}_{1}$ $\mathrm{c}=1 \Rightarrow \mathrm{a}+\mathrm{b}=20 \rightarrow{ }^{20+2-1} \mathrm{C}_{2-1}={ }^{21} \mathrm{C}_{1}$ $\vdots \vdots \vdots$ $\mathrm{c}=10 \Rightarrow \mathrm{a}+\mathrm{b}=2 \rightarrow{ }^{2+2-1} \mathrm{C}_{2-1}={ }^{3} \mathrm{C}_{1}$ $\mathrm{c}=11 \Rightarrow \mathrm{a}+\mathrm{b}=0 \rightarrow{ }^{0+2-1} \mathrm{C}_{2-1}={ }^{1} \mathrm{C}_{1}$ Number of non-negative integral solutions ${ }^{1} \mathrm{C}_{1}+{ }^{3} \mathrm{C}_{1}+{ }^{5} \mathrm{C}_{1}+\ldots+{ }^{23} \mathrm{C}_{1}$ $1+3+5 \ldots \ldots+23$ $=144$

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.

Solve a similar one free

Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Permutations and Combinations
Topic: Number of integral solution of linear Equations