JEE Main 2026 — Permutation Combination Question with Solution

The given equation is $a+b+2c=22$, where $a,b,c$ are non-negative integers.

This can be rewritten as $a+b=22-2c$.

Since $a\ge 0$ and $b\ge 0$, we must have $22-2c\ge 0$, which gives $c\le 11$.

Since $c$ is a non-negative integer, the possible values for $c$ are $0,1,2,\dots ,11$.

For a fixed value of $c$, the number of non-negative integer solutions to the equation $a+b=22-2c$ is given by $(22-2c)+1=23-2c$.

The total number of solutions $n(A)$ is the sum of the number of solutions for each possible value of $c$:

$n(A)={\sum }_{c=0}^{11}(23-2c)$

$n(A)=23+21+19+\cdots +1$

This is an arithmetic progression with $12$ terms, first term $1$, and last term $23$.

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

Answer: $144$