What I naturally did was to find the prime factorisation of the product of factorials which is $ 2^{30}3^{13}5^5 7^3 $. Clearly there is 15 unique perfect squares that divide $2^{30}$, 6 unique perfect squares that divide $3^{13}$, 2 unique perfect squares that divide $ 5^5$ and 1 unique perfect square that divides $7^3$. This gives us a total of 24 perfect squares so far. Now we create a set $ A$ of 24 unique elements, where each element represents one of the 24 unique perfect squares. (Note that we have not counted $ 1^2$ as a perfect square that divides the product yet; we'll do this later).
$ S=\{p_1,p_2,p_3,p_4,…,p_{24}\} $
Since we have that the product of any number of perfect squares is also a perfect square, then we have count to the number of ways we can choose $1,2,3,4,…,23,24$ elements from the set since the product of $n$ elements, where $ 1 \le n \le 24$, is also a perfect square. Hence we evaluate:
$ \displaystyle \sum_{n=1}^{24} {24 \choose n}=2^{24}-1 $
Since we still have to count $ 1^2$, we add that in hence the number of perfect squares that divide $ \displaystyle \prod_{n=1}^9n!$ is $ 2^{24}-1+1=\boxed{2^{24}} $.
However this answer is incorrect since the answer is supposed to be an integer in the range $[0,999]$ (inclusive) but my answer is clearly way off. I don't know where I made a mistake and so I'd appreciate any help. Thanks.
Best Answer
The perfect square can be represented as $$2^a3^b5^c7^d$$ where each of $a,b,c,d$ is even.
For example, $a$ has $16$ possibilities as $a=0,2,4,\cdots, 28, 30.$
Since you can choose any even in each letter, the answer will be $$16\times 7\times 3\times 2=672.$$
In general, the number of the perfect square which divide $${p_1}^{a_1}{p_2}^{a_2}\cdots {p_k}^{a_k}$$ will be $$\left({\left\lfloor \frac{a_1}{2}\right\rfloor}+1\right) \left({\left\lfloor \frac{a_2}{2}\right\rfloor}+1\right) \cdots \left({\left\lfloor \frac{a_k}{2}\right\rfloor}+1\right).$$
Here, $p_1\lt p_2\lt\cdots\lt p_k$ are prime numbers.