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?

Two minutes is a lot of calculations, I'd write the 19 numbers in the numerator and the 19 numbers in the denominator, and cancel whatever can be cancelled in under a minute.
You get:
$$ 3^3\times5^2\times7^2\times23\times29\times31\times47\times53\times59\times61$$
We approximate this as:
$$20\times 20 \times 50 \times 20 \times 20 \times 20 \times 50 \times 50\times 50 \times 50=10^{15}$$
The actual value is $6.131164307078475\times 10^{15}$
