Maths Q17 – JEE-MAIN 2026

Home Maths Permutations and CombinationsQuestion #317

Let

A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2 c = 22}

. Then

n (A)

is equal to:

Clue 1 of 3: Combinatorics and Diophantine Equations

The problem asks for the number of non-negative integer solutions to a linear equation, which is a classic combinatorics problem.

🧠 Full Solution Path

Step 1: Determine the range of 'c'✦ Active

The given equation is $a + b + 2 c = 22$ , where $a, b, c$ are non-negative integers. Since $a \geq 0$ and $b \geq 0$ , we must have $2 c \leq 22$ . This implies $c \leq 11$ . Therefore, $c$ can take any integer value from $0$ to $11$ .

Step 2: Calculate solutions for 'a+b' for each 'c'○ Expand

For each value of $c$ , the equation reduces to $a + b = 22 - 2 c$ . Let $K = 22 - 2 c$ . The number of non-negative integer solutions for $a + b = K$ is $K + 1$ . We iterate through $c$ :

\begin{array}{ll} c = 0 : & a + b = 22 ⟹ 22 + 1 = 23 solutions \\ c = 1 : & a + b = 20 ⟹ 20 + 1 = 21 solutions \\ c = 2 : & a + b = 18 ⟹ 18 + 1 = 19 solutions \\ ⋮ & ⋮ \\ c = 11 : & a + b = 0 ⟹ 0 + 1 = 1 solution \end{array}

💡 Teacher's Secret Hint

Remember that the number of non-negative integer solutions to $x_{1} + x_{2} = N$ is $N + 1$ .

Step 3: Sum the number of solutions○ Expand

The total number of solutions $n (A)$ is the sum of the arithmetic series: $23 + 21 + 19 + \dots + 1$ . This series has $12$ terms (from $c = 0$ to $c = 11$ ). The sum of an arithmetic series is given by $S_{n} = \frac{n}{2} (f i r s t_t e r m + l a s t_t e r m)$ .

n (A) = \frac{12}{2} (23 + 1) = 6 (24) = 144

💡 Teacher's Secret Hint

Ensure you correctly identify the number of terms in the arithmetic progression.

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Maths Question 17 – JEE-MAIN 2026

Clue 1 of 3: Combinatorics and Diophantine Equations

🧠 Full Solution Path