Question

Three natural numbers are taken at random from the set $A=\{x:1\le x\le 100,x\in N\}$. The probability that $A.M.$ of the numbers taken is $25$ is

• A. $\frac{{}^{77}{C}_2}{{}^{100}{C}_3}$
• B. $\frac{{}^{25}{C}_2}{{}^{100}{C}_3}$
• C. $\frac{{}^{74}{C}_2}{{}^{100}{C}_3}$
• D. None of these

Answer 1

Answer: (C)

$\frac{{}^{74}{C}_2}{{}^{100}{C}_3}$

$n\left(S\right){=}^{100}{C}_3$
Since the $A.M.$ of three numbers is $25$.
$\therefore$ Their sum $=75$
$\therefore n\left(E\right)=$ the number of integral solutions of $x_1+x_2+x_3=75$
where $x_1\ge 1,x_2\ge 1,x_3\ge 1.$
$=$ coefficient of $x^{7}5$ in ${(x+x^2+x^3+...)}^3$
$=$ coefficient of $x^{7}2$ in ${(1+x+x^2+x^3+...)}^3$
$=$ coefficient of $x^{7}2$ in ${\frac{(}{1}1-x)}^3$
$=$ coefficient of $x^{7}2$ in ${(1-x)}^{-3}$
${=}^{74}{C}_{72}{=}^{74}{C}_2$
$\therefore P\left(E\right)=\frac{{}^{74}{C}_2}{{}^{100}{C}_3}$