Can I apply stars and bars here? Let $K$ be the number of ordered tuples $(a, b, c, d, e)$ of positive integers such that $abcde = 69!$ . How many positive divisors does the number of positive divisors of $K^3$ have?
Can I use stars and bars on this? If so, how can I use that to solve this?
 A: When $p≤n$ is a prime then there are
$$\sum_{k\geq1}\left\lfloor{n\over p^k}\right\rfloor$$
factors $p$ in $n!\>$. This gives you
$$69!=2^{66}\ 3^{32}\ 5^{15}\ 7^{10}\ 11^6\ 13^5\ 17^4\ 19^3\ 23^3\ 29^2\ 31^2\ 37\ 41\ \ldots 67\ .$$
The $66$ factors $2$ can be distributed in ${66+4\choose4}$ ways into the five pots $a$, $\ldots$, $e$ (that's "stars and bars"), and similarly for the other prime factors. This gives
$$K={70\choose4}\ {36\choose 4}\ \ldots \ {7\choose 4}^2\ {6\choose 4}^2\cdot 5^8\ .$$
We don't need the decimal representation of $K$, but its prime factorization. The latter can be found by hand (haha) from the above binomials, and is
$$K=2^5\ 3^8\ 5^{16}\ 7^8\ 11^2\ 13^1\ 17^3\ 19^1\ 23^1\ 67^1\ .$$
This at once gives the prime factorization of $K^3$.
When $K^3=\prod_j p_j^{\alpha_j}$ then $K^3$ has
$$\prod_j(\alpha_j+1)$$
positive divisors. I obtained $8\,780\,800\,000$. Now you have to find the number of positive divisors of this. For that we need
$$8\,780\,800\,000=2^{13}\ 5^5\ 7^3\ ,$$
so that the end result seems to be $336$.
