There are three piles of identical yellow, black and green balls and each pile contains at least $20$ balls.
The number of ways of selecting $20$ balls if the number of black balls to be selected is twice the number of yellow
balls, is

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answer is B.

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Detailed Solution

Let the number of yellow balls be $x,$ that of
black be $2 x$ and that of green be $y .$ Then

$\begin{aligned} x + 2 x + y = 20 or 3 x + y = 20 \Rightarrow^{} y = 20 - 3 x \end{aligned}$

As $0 \leq y \leq 20, we get 0 \leq 20 - 3 x \leq 20$

$\Rightarrow 0 \leq 3 x \leq 20 or 0 \leq x \leq 6$

$∴$ The number of ways of selecting the balls is $7 .$

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