Question

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

A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these

Solution

## The correct options are B $$\displaystyle \dfrac{^{89}\textrm{C}_{2}}{^{50}\textrm{C}_{47}}$$ C $$\displaystyle \dfrac{^{89}\textrm{C}_{87}}{^{50}\textrm{C}_{3}}$$$$n(S) =^{50}\textrm{C}_{3}$$As average value given $$30$$, mean sum of the numbers $$=90$$$$\therefore n(A) =$$ 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 $$\left ( x+x^{2}+x^{3}\cdots \right )^{3}$$$$=$$ Coefficient of $$x^{87}$$ in $$\left ( 1+x+x^{2}+\cdots \right )^{3}$$$$=$$ Coefficient of $$x^{87}$$ in $$\left ( 1-x \right )^{-3}$$$$=$$ Coefficient of $$x^{87}$$ in $$\left ( 1+ ^{3}\textrm{C}_{1}x+^{4}\textrm{C}_{2}x^{2}+\cdots +^{89}\textrm{C}_{87}x^{87}+\cdots \right )$$$$=^{89}\textrm{C}_{87}=^{89}\textrm{C}_{2}$$$$\therefore$$Required probability $$=\displaystyle \dfrac{^{89}\textrm{C}_{87}}{^{50}\textrm{C}_{3}}=\dfrac{^{89}\textrm{C}_{2}}{^{50}\textrm{C}_{47}}$$Maths

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