Stirlings Approximation : $n! \approx \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}$.
So $100!$ has an approximate percentage error of about $\frac{100}{12n} = \frac{1}{12}$. Using this information, how does one go about finding the percentage error of using Stirlings to calculate:
$$ \frac{100!}{50!50!} $$
In other words what is the normal way to adjust the approximation error, in this example I am using stirlings 3 times
You have:
$$\mathrm e^{1/(12n+1)} \leq \frac{n!}{\sqrt{2\pi n}\cdot(\frac n{\mathrm e})^n} \leq \mathrm e^{1/(12n)}$$
Let $f(n)$ be $\frac{n!}{\sqrt{2\pi n}\cdot(\frac n{\mathrm e})^n}$. You have:
$$\mathrm e^{1/(12n+1)} \leq f(n) \leq \mathrm e^{1/(12n)}$$
and
$$\mathrm e^{-1/(12n)} \leq \frac{1}{f(n)} \leq \mathrm e^{-1/(12n+1)}$$
Thus:
$$\mathrm e^{1/(12\cdot100+1)} \cdot \mathrm e^{-2/(12\cdot50)} \leq \frac{f(100)}{f(50)f(50)} \leq \mathrm e^{1/(12\cdot100)} \cdot \mathrm e^{-2/(12\cdot50+1)}$$