How to estimate the size of a ratio with very large factorials?

Question

I want to estimate the size of the following ratio: $$\frac{10^{18}!}{10^{14}!\ 10^4!}$$

Since I don't have an idea how to simplify it and no CAS is able to handle numbers of this size, I am at an impasse right now.

Answer 1

By Stirling's Approximation, $ln(10^{14}!) \sim 10^{14}ln(10^{14})-10^{14} \sim 14 \times 10^{14}ln(10)- 10^{14} \sim 10^{15}$

$ln(10^{18}!) \sim 10^{18}ln(10^{18})-10^{18} \sim 18 \times 10^{18}ln(10)- 10^{18} \sim 18 \times 10^{18}ln(10)$

$ln(10^{4}!) \sim 10^{4}ln(10^{4})-10^{4} \sim 4 \times 10^{4}ln(10)- 10^{4} \sim 10^{4}$ $\\$

$ln(\frac{10^{18}!}{10^{14}!{10^4}!})\sim 18 \times 10^{18}ln(10)-10^{15}-10^{4} \sim 18 \times 10^{18}ln(10)$

$\frac{10^{18}!}{10^{14}!{10^4}!} \sim 10^{18 \times 10^{18}}$

Answer 2

I think you are going to need Stirling's approximation to make much headway here:

$$n! \sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n},$$

so that $$\log(n!) \sim n \log n - n + \frac12\log(2\pi n).$$

Using that, you can get $$\log\left(\frac{10^{18}!}{10^{14}!10^4!}\right) = 10^{18}\log(10^{18}) - 10^{18} +\frac12\log(2\pi\times 10^{18}) - 10^{14}\log(10^{14}) + \cdots$$

and then discard whichever terms are too small to be worth bothering with in your desired application.