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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
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B
89C250C47
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C
89C8750C3
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D
None of these
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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 $$=\displaystyle
\dfrac{^{89}\textrm{C}_{87}}{^{50}\textrm{C}_{3}}=\dfrac{^{89}\textrm{C}_{2}}{^{50}\textrm{C}_{47}}$$

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