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

Let the number of green balls be x. Then, the number of red balls is 2x. Let the number of blue balls bey. 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.