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

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 $c$:

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

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

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

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

Answer: $144$