JEE Main 2026 — Mathematics Algebra

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

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

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

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

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

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

$$n(A) = \sum_{c=0}^{11} (23 - 2c)$$n(A)=c=0∑11​(23−2c)

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

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

$$n(A) = \dfrac{12}{2} (1 + 23) = 6 \times 24 = 144$$n(A)=212​(1+23)=6×24=144

Answer: $144$144