How many triples of positive integers satisfy $x * y * z = 9 * 2^{20}$? How many triples of positive integers satisfy $x * y * z = 9 * 2^{20}$? 

The order in which the numbers are considered does not matter.

I figure the total number of permutations is $9(\frac{22!}{20!}2!)$.

I tried to solve it by first considering all the triples with two equal numbers such as $2^0 * 2^0 * 9$... up to $2^{10} * 2^{10} * 9$, for a total of $11$ classes (and $11 * 3$ if we consider the order).

Then I figured that since $9$ can be written as $3 * 3 * 1$, I should also count in triples like $(2^0 * 3) * (2^0 * 3) * 2^{20}$, and there are other 11 of these.

And lastly I calculated the following : (Tot - $66$) : $6$, and added $22$.
However, it didn't work so well. What am I getting wrong? (probably more than one thing)
Hint: The product must contain 2 factors of 3, and 20 factors of 2.  This means that you have to choose where to put the two 3's (i.e. $x$ contains 0, 1 or 2, $y$ contains 0, 1 or 2, $z$ contains 0, 1 or 2) and where to put the twenty 2's.  The number of ways of doing that is the same as the number of ways to write 20 (or 2 respectively) as the sum of 3 non-negative integer variables.  And that question has been answered a number of times here.
The answer is $\binom{22-1}{3-1} = 210$. 
Its a classic combinatorics (stars and bars) problem.