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Out of 5 apples, 10 mangoes and 15 oranges, any 15 fruits distributed among two persons. The total number of ways of distribution
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The correct answer is:
$66$
Number of ways $=$ coefficient of $x^{15}$ in the expansion $\left(1+x+x^2+x^3+x^4+x^5\right)$
$\left(1+x+x^2+\ldots \ldots . . x^{10}\right)$
$\left(1+x+x^2+\ldots \ldots+x^{15}\right)$
$\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x+x^2+\ldots .+x^{10}\right)$
$\left(1+x+x^2+\ldots+x^{15}\right)=\left(1-x^6-x^{11}\right)\left(1+{ }^3 C_1 x+{ }^4 C_2 x^2\right.$
$\left.+\ldots \ldots+{ }^6 C_4 x^4+{ }^{11} C_9 x^9+{ }^{17} C_{15} x^{15}+\ldots \ldots \ldots . ..\right)$
$=\ldots \ldots . .+\ldots \ldots .+x{ }^{15}\left(-{ }^{11} C_9-{ }^6 C_4+{ }^{17} C_{15}\right)$
$=\ldots \ldots . .+\ldots \ldots+x^{15}(-55-15+136)=x^{15} \times 66$
$\therefore$ Coefficient of $x^{15}=66$.
$\left(1+x+x^2+\ldots \ldots . . x^{10}\right)$
$\left(1+x+x^2+\ldots \ldots+x^{15}\right)$
$\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x+x^2+\ldots .+x^{10}\right)$
$\left(1+x+x^2+\ldots+x^{15}\right)=\left(1-x^6-x^{11}\right)\left(1+{ }^3 C_1 x+{ }^4 C_2 x^2\right.$
$\left.+\ldots \ldots+{ }^6 C_4 x^4+{ }^{11} C_9 x^9+{ }^{17} C_{15} x^{15}+\ldots \ldots \ldots . ..\right)$
$=\ldots \ldots . .+\ldots \ldots .+x{ }^{15}\left(-{ }^{11} C_9-{ }^6 C_4+{ }^{17} C_{15}\right)$
$=\ldots \ldots . .+\ldots \ldots+x^{15}(-55-15+136)=x^{15} \times 66$
$\therefore$ Coefficient of $x^{15}=66$.
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