Question

There are three piles of identical red, blue and green balls and each pile contains at least $10$ balls. The number of ways of selecting $10$ balls if twice as many red balls as green balls are to be selected, is

• A. $3$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is B $4$

Let the number of green balls be $x$. Let the number of red balls is $2x$. Let the number of blue balls be $y$. Then, $x+2x+y=10\Rightarrow 3x+y=10\Rightarrow y=10-3x$

Clearly, $x$ can take values $0,1,2,3$. The corresponding values of $y$ are $10,7,4$ and $1$. Thus the possibilities are $(0,10,0),(2,7,1),(4,4,2)$ and $(6,1,3)$, where $(r,b,g)$ denotes the number of red, blue and green balls