<s>[INST] You have to estimate $\binom{63}{19}$ in $2$ minutes to save your life. This is from the lecture notes in this course of discrete mathematics I am following. 
The professor is writing about how fast binomial coefficients grow.

So, suppose you had 2 minutes to save your life and had to estimate, up to a factor of $100$, the value of, say, $\binom{63}{19}$. How would you do it? I will leave this (hopefully intriguing!) question hanging and maybe come back to the topic of efficiently estimating binomial coefficients later.

Any ideas/hints on how to do it?
 [/INST] With pen and paper, Stirling's approximation:
$$
\begin{align}
{63 \choose 19}
&= \frac{63!}{19! 44!} \\
&\doteq \frac{\sqrt{2\pi 63}}{\sqrt{2\pi 19}\sqrt{2\pi 44}}
        \left( \frac{63}{e}\right)^{63}
        \left( \frac{e}{19}\right)^{19}
        \left( \frac{e}{44}\right)^{44} \\
&\doteq \sqrt{\frac{60}{2 \cdot 3 \cdot 20 \cdot 50}} \cdot
        20^{63} \cdot 5^{-19} \cdot 15^{-44} \\
&\doteq \frac{1}{10} \cdot 2^{63} \cdot 10^{63} \cdot 2^{19} \cdot 10^{-19}
                     \cdot 3^{-44} \cdot 2^{44} \cdot 10^{-44} \\
&= 10^{-1} \cdot 2^{126} \cdot 3^{-44} \\
&= 10^{-1} \cdot (2^{10})^{12} \cdot 2^6 \cdot (3^2)^{-22} \\
&\doteq 50 \cdot 10^{-1+36-22} \\
&= 5 \cdot 10^{14}
\end{align}
$$
using various estimates including $2^{10} \doteq 10^3$.
Crude, but if instead you had to estimate ${630 \choose 19}$ then you might not want to do cancelling by writing out integer factors.
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