Ball and urn method (counting problems) 
How many ordered triples $(a, b, c)$ of positive integers exist with the property that $abc = 500$?

Since, $500 = 2^2 5^3$
I believe this can be solved using Ball and Urn
let $a = 2^{x_1}5^{y_1}$ and then the subscripts increase by $1$ for $b, c$. 
There is:
$$x_1 + x_2 + x_3 = 2$$
$$y_1 + y_2 + y_3 = 3$$
We count each seperately. 
There are 3 "bins," $x_1, x_2, x_3$. 
Then what are the objects? Im confused?
 A: I think this is the same as the ball and urn method:
We can distribute the two factors of $2$ among $a,b,c$ using the stars and bars method: each arrangement corresponds to an arrangement of $\{\star,\star,|,|\}$. There are $\binom{4}{2}$ ways to arrange these.
We can distribute the three factors of $5$ among $a,b,c$ using the stars and bars method: each arrangement corresponds to an arrangement of $\{\star,\star,\star,|,|\}$. There are $\binom{5}{2}$ ways to arrange these.
A: Expanding on robjohn's answer.
The 'bins' (or urns) are the factors $a, b, c$.
The number of 'balls' are counted by the exponents of $2$. Since $500= 2^25^3$ there are two 'balls' to distribute among the three bins ($a,b,c$) to determine what exponent of $2$ each might receive.
$$\begin{array}{lll}a & b & c \\ \hline
2^2 & 2^0 & 2^0 & \ast\ast \vert\vert
\\ 2^1 & 2^1 & 2^0  & \ast\vert\ast \vert
\\ 2^1 & 2^0 & 2^1 & \ast\vert\vert\ast
\\ 2^0 & 2^2 & 2^0 & \vert\ast\ast\vert
\\ 2^0 & 2^1 & 2^1 & \vert\ast\vert\ast
\\ 2^0 & 2^0 & 2^2 & \vert\vert\ast\ast
\end{array}\qquad \dbinom{2+2}{2,2}=\dfrac{4!}{2!2!} 6$$
And similarly we count the ways the exponents of $5$ may be assigned by counting the permutations of $3$ stars and $2$ bars.$$\dbinom{3+2}{3,2} = \dfrac{5!}{3!2!}=10$$
