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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Let $x$ denote the number of ways of arranging $m$ boys and $m$ girls in a row so that no two boys sit together. If $y$ and $z$ denote the number of ways of arranging $m$ boys and $m$ girls in a row and around a circular table respectively so that boys and girls sit alternately,then $x: y: z=$

$^{37}C_4 + \sum_{r=1}^{5} {^{(42-r)}C_r} = $

If $n$ is an integer with $0 \leq n \leq 11$,then the minimum value of $n!(11-n)!$ is attained when a value of $n$ equals to

The value of ${}^6P_4 + 4 \cdot {}^6P_3$ is $.......$ .

If $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$,find $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module