Given that $xyz = 2^{15} * 3^{20}$ we know that $x = 2^{a}3^{j}, y = 2^{b}3^{k}$ and $z = 2^{c}3^{l}$ where

• (i) $a + b + c = 15$ , where $a,b,c$ are all non-negative integers, and

• (ii) $j + k + l = 20$ , where $j,k,l$ are also all non-negative integers.

The number of solutions to each of these equations can be determined by the stars and bars method . Equation (i) then has $\dbinom{15 + 3 - 1}{15} = \dbinom{17}{15}$ solutions and equation (ii) has $\dbinom{20 + 3 - 1}{20} = \dbinom{22}{20}$ solutions.

The desired number of ordered triples $(x,y,z)$ is then the product of these two values, namely

$\dbinom{17}{15} * \dbinom{22}{20} = 136*220 = \boxed{31416}$ .

Suppose x ,y and z have a ,b and c power of 2 contained in them respectively.

hence a+b+c=15 where a,b,c>=0 . Its no. Of solutions is 17C2 .

Similarily doing for 3 no. Of solutions is

22C2.

Hence the answer is

17C2 *22C2 = 31416