How many ordered quadruples $(w, x, y, z)$ of non-negative integers are there such that $wxyz = 288$? Why? I found the prime factorization
$$288 = 2^5 \cdot 3^2$$
and then I tried to set up some algebraic equations, but got stuck. How would we proceed so that we get the answer? Especially when we are only looking at non-negative integers. Any help?
 A: Forget about algebraic equations.
You just need to create $4$ integers using these $7$ prime factors: $2,2,2,2,2,3,3$
So, for example, you can take two $2$'s, another two $2$'s,  a $2$ and a $3$, and finally just a single $3$, resulting in $(4,4,6,3)$
Now, one important observation: any of the numbers could be a $1$ ... basically meaning that you wouldn't pick any of the $7$ prime factors for that one.
So, the question is: what other (and how many) ways are there to create $4$ numbers using these $7$ prime factors?
Fortunately, since the numbers are ordered, you can use start and bars. First, divvy up the $2$'s:
E.g.
$2||22|22$ corresponds to $w$ getting a factor of $2$, $x$ not geting any $2$, $y$ getting two $2$'s and $z$ getting two $2$'s
With 5 'stars and 3 'bars', where the bars can go in any of the 8 spots, that's $8 \choose 3$ possibilities there.
Now divvy up the $3$'s ... so now you have 2 'stars and 3 'bars', so there are $5 \choose 3$ possibilities there.
Multiply, and you get the number of all possible combinations:
$${8 \choose 3}{5 \choose 3}$$
A: Each of the variables is of the form $2^{a_i}\cdot 3^{b_i}$ where:
$$ a_1+a_2+a_3+a_4 =5$$ and $$ b_1+b_2+b_3+b_4 =2$$
and $a_i,b_j$ are nonnegative integers. By stars and bars method we have $${8\choose 3}\cdot {5\choose 3}= 56\cdot 10 = 560$$ solutions.
