Question

Three natural numbers are taken as random from a set of numbers $\left\{1,\mathrm{2....50}\right\}$. The probability that their average value taken as $30$ equals

• A. $\frac{{}^{30}{C}_2}{{}^{89}{C}_2}$
• B. $\frac{{}^{89}{C}_2}{{}^{50}{C}_{47}}$
• C. $\frac{{}^{89}{C}_{87}}{{}^{50}{C}_3}$
• D. None of these

Answer 1

Answer: (B), (C)

$n\left(S\right)={}^{50}{C}_3$
As average value given $30$, mean sum of the numbers $=90$
$n\left(A\right)$ = The number of solution of the equation $x_1+x_2+x_3=90$
$\Rightarrow x_1,x_2,x_3,$ each greater than $1$.
= Coefficient of $x^{90}$ in ${\left(x+x^2+x^3+...\right)}^3$
= Coefficient of $x^{87}$ in ${\left(1+x+x^2+...\right)}^3$
= Coefficient of $x^{87}$ in ${\left(1-x\right)}^{-3}$
= Coefficient of $x^{87}$ in $\left(1+{}^3{C}_{1}x+{}^4{C}_2{x}^2+...+{}^{89}{C}_{87}{x}^{87}+...\right)$
$={}^{89}{C}_{87}={}^{89}{C}_2$
$\therefore$ Required probability$=\frac{{}^{89}{C}_{87}}{{}^{50}{C}_3}=\frac{{}^{89}{C}_2}{{}^{50}{C}_{47}}$