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?
 A: 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 $6.131164307078475\times 10^{15}$
A: Use averages between two equally displaced factors: $\frac {63 \times  62 ... \times 54 ...\times 45}{19 \times 18 \times ... \times 9 ...\times 1} \approx  \frac {54^{19}}{9^{19}} = 6^{19}$
Based on formula $(N+n)\times(N-n)= N^2-n^2$, $n^2$ "small" to $N^2$,
which is good enough for numerator and not so good for denominator, so because of easier calculation writing 9 instead of 10.
$6^{19} = 6.09359740010496\times 10^{15}$, well it is factor of 10, but almost exact and most important, really achievable within two minutes.
A: 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.
A: So we want to estimate:
$$\frac{63!}{19!44!} $$
We do this by doing a geometric average approximation, replacing each factorial by it's geometric average, and the following approximations $9.5^{19} \approx 9^{20}$ and since it's a matter of life and death we expose we know the 10% and 5% interest tables by heart (useful since $22 = 20\cdot 1.10$ and $31.5 = 30\cdot 1.05$ so the exponents will copy over to the "percentage" part):
$$\frac{31.5^{63}}{ 9.5^{19} \cdot 22^{44}} = \frac{22\cdot 3^{63}\cdot10^{63}}{66\cdot 2^{44}\cdot10^{44}\cdot 3^{2\cdot 20}} = \frac{1}{3}\frac{3^{23}10^{19}}{2^{44}}$$ then cancelling by using $2^3 \approx 3^2 \approx 10$: 
$$ \approx \frac{10^{19}}{2^{11}} \approx 2.5\cdot 10^{15}$$
A: In the numerator we have $63!/(63-19)!\approx(63-9)^{19}=54^{19}\approx50^{20}=100^{20}/2^{20}\approx10^{34}$.
In the denominator we have $19!\approx\left(\frac{1+19}2\right)^{19}=10^{19}$.
So the quotient is roughly $10^{15}$.
I'm not sure I could have done that in two minutes under the threat of death, though.
A: Here you go!
${63\choose19} = 6.13116E+15$
${63\choose19} = \frac{63!}{19!44!}$ $$ = \frac{63^{63}}{19^{19}.44^{44}}$$$$ \approx \frac{60^{60}}{20^{20}.40^{40}}$$$$\approx \frac{3^{60}}{2^{40}}$$$$\approx 16.3^{32}$$$$\approx  9^{16}$$$$\approx 10^{16} $$
This uses Sterling's approximation for factorial (n)
