Question

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

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

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{{}^{89}{\text{C}}_2}{{}^{50}{\text{C}}_{47}}$
C $\frac{{}^{89}{\text{C}}_{87}}{{}^{50}{\text{C}}_3}$

$n\left(S\right){=}^{50}{\text{C}}_3$
As average value given $30$, mean sum of the numbers $=90$
$\therefore n\left(A\right)=$ 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 ${(x+x^2+x^3\cdots )}^3$
$=$ Coefficient of $x^{87}$ in ${(1+x+x^2+\cdots )}^3$
$=$ Coefficient of $x^{87}$ in ${(1-x)}^{-3}$
$=$ Coefficient of $x^{87}$ in $(1{+}^3{\text{C}}_{1}x{+}^4{\text{C}}_2{x}^2+\cdots {+}^{89}{\text{C}}_{87}{x}^{87}+\cdots )$
${=}^{89}{\text{C}}_{87}{=}^{89}{\text{C}}_2$
$\therefore$Required probability $=\frac{{}^{89}{\text{C}}_{87}}{{}^{50}{\text{C}}_3}=\frac{{}^{89}{\text{C}}_2}{{}^{50}{\text{C}}_{47}}$