Question

Three natural numbers are taken as random from a set of numbers $\left\{1, 2 ....50\right\}$. 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$= \frac{\,{}^{89}C_{87}}{\,{}^{50}C_{3}} = \frac{\,{}^{89}C_{2}}{\,{}^{50}C_{47}}$