Question

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 thrice the number of yellow balls is?

• A. $6$
• B. $7$
• C. $8$
• D. $9$

Answer 1

The correct option is B

$7$

Explanation of the correct answer

Let, $a$ is the number of yellow balls , $b$ is the number of black balls and $2a$ is the number of green balls.

Given : Each pile contains at least $20$ balls.

$\begin{array}{lcl}a+2a+b& =& 20\\ \Rightarrow 3a+b& =& 20\\ \Rightarrow b& =& 20-3a\end{array}$

Since $b$ lies between $0$ and $20$ and $a$$,b$ will be a integer.

$\begin{array}{ccccc}0& \le & b& \le & 20\\ 0& \le & 20-3a& \le & 20\\ -20& \le & -3a& \le & 0\\ 0& \le & 3a& \le & 20\\ 0& \le & a& \le & 6\end{array}$

Therefore, ways of selecting yellow balls is $7$.

Hence option $\left(B\right)$, $7$ is the correct answer.