Count Non Negative Integer Solutions for Linear Equation

To find the cardinality $n(A)$ of the set $A = \{(a, b, c) : a, b, c \in \mathbb{Z}_{\ge 0} \text{ and } a + b + 2c = 22\}$, we need to count the total number of non-negative integer solutions to the equation:

$a + b + 2c = 22$

We can express $a + b$ in terms of $c$:

$a + b = 22 - 2c$

Since $a \ge 0$ and $b \ge 0$, their sum must be non-negative:

$22 - 2c \ge 0 \implies c \le 11$

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

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

$(22 - 2c) + 1 = 23 - 2c$

To find $n(A)$, we sum the number of solutions over all possible non-negative integer values of $c$ from $0$ to $11$:

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

Expanding the sum:

• For $c = 0$: $23 - 0 = 23$
• For $c = 1$: $23 - 2 = 21$
• For $c = 2$: $23 - 4 = 19$
• $\vdots$
• For $c = 11$: $23 - 22 = 1$

Thus, $n(A)$ is the sum of the first $12$ odd integers:

$n(A) = 23 + 21 + 19 + \dots + 3 + 1$

Using the formula for the sum of the first $N$ odd numbers, $\sum_{k=1}^{N} (2k - 1) = N^2$, where $N = 12$:

$n(A) = 12^2 = 144$

Hence, $n(A)$ is equal to $144$, which corresponds to Option C.