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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.

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May be you can use Stirling's approximation for the factorials. –  Pratyush Sarkar Oct 6 '13 at 15:22
    
I think this is a good starting point but perhaps there are other ideas?!? –  vonjd Oct 6 '13 at 15:27
    
What is the source of this ratio? It's 'binomial-like' but it's not, e.g., the binomial coefficient $10^{18}\choose 10^{14}$, which would make a measure of sense; instead it's essentially $\frac{(ab)!}{(a!)(b!)}$ which doesn't really have any clear combinatorial interpretation. –  Steven Stadnicki Oct 7 '13 at 4:26
    
@StevenStadnicki: You are right, I made a stupid beginner's mistake :-( I wanted to estimate $\frac{10^{18}!}{10^{14}!\ (10^{18}-10^{14})!}$ –  vonjd Oct 8 '13 at 13:07
    
@vonjd Assuming it's not too late for whatever reason, I suggest posting that question up independently. I also encourage leaving it as a binomial coefficient rather than breaking it down; there are well-established asymptotic estimates for binomial coefficients that can be applied to the problem. (Possibly use a title something like 'estimating the size of very large binomial coefficients'). –  Steven Stadnicki Oct 8 '13 at 15:46

2 Answers 2

up vote 2 down vote accepted

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

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

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

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

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

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I tried it in WolframAlpha and it gave me an even bigger result: The highest exponent might even be greater then 19 (which would make a huge difference!) - Input: ((10^18)/e)^(10^18)*Sqrt(2*Pi*(10^18))/(((10^14)/e)^(10^14)*Sqrt(2*Pi*(10^14))*(‌​(10^4)/e)^(10^4)*Sqrt(2*Pi*(10^4))) –  vonjd Oct 6 '13 at 15:46
    
Oh i'm sorry, I made a careless mistake. I just edited the answer, so that the new answer becomes k^18, where the old answer is k. So, this is a huge difference. –  Aran Komatsuzaki Oct 7 '13 at 4:21

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.

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