Let A = {(a, b, c) : a, b, c text{ are non-ne | vedclass

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(C) The given equation is $a + b + 2c = 22$, where $a, b, c \ge 0$.We can rewrite this as $a + b = 22 - 2c$.Since $a, b \ge 0$, we must have $22 - 2c

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(C) The given equation is $a + b + 2c = 22$, where $a, b, c \ge 0$.We can rewrite this as $a + b = 22 - 2c$.Since $a, b \ge 0$, we must have $22 - 2c \ge 0$, which implies $0 \le c \le 11$.For a fixed $c$, the number of non-negative integer solutions to $a + b = 22 - 2c$ is given by the formula $(n+r-1)C(r-1)$, which simplifies to $(22 - 2c + 1) = 23 - 2c$.Thus, the total number of solutions is $n(A) = \sum_{c=0}^{11} (23 - 2c)$.This is an arithmetic progression: $n(A) = 23 + 21 + 19 + \dots + 1$.The number of terms is $12$.The sum is $\frac{12}{2} 	imes (23 + 1) = 6 	imes 24 = 144$.

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

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