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How can we compare combination with power when the both are very very large numbers?

For example how to compare $\binom{1000}{500}$ with $2^{500} , 3^{500} , 4^{500}$ ?

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    You could use logarithms and [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation) for factorial. – 2017-01-04
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    For $4^{500}=2^{1000}=(1+1)^{1000}=^{1000}C_0+^{1000}C_1...+^{1000}C_{500}+...+^{1000}C_{1000}$. Hence $^{1000}C_{500}<4^{500}$ – 2017-01-04

3 Answers 3

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First, note that if $\log(a) > \log(b)$, then $a > b$.

Stirlings approximation lets us write $$\log(n!) = n\log(n)-n$$ So that since $${1000 \choose 500} = \frac{1000!}{500! 500!}$$ we have $$\log\left(\frac{1000!}{500! 500!}\right) = \log(1000!) - \log((500!)^2)\approx (1000\log(1000) - 1000) - 2(500\log(500) - 500) = 1000(\log(1000)-\log(500)) = 1000 \log(2) $$ we can compare this to, for example, $2^{500}$ by comparing their logarithms: $$\log\left(\frac{1000!}{500! 500!}\right) \approx 1000 \log(2) > \log(2^{500}) = 500 \log(2)$$ $$1000 \log(2) > 500 \log(2)$$ so that ${1000 \choose 500} > 2^{500}$.

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Partial answer:

$4^{500}=(2^2)^{500}=2^{2\cdot500}=2^{1000}=\sum\limits_{n=0}^{1000}\binom{1000}{n}>\sum\limits_{n=500}^{500}\binom{1000}{n}=\binom{1000}{500}$

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Using the definition of choose, you could write \begin{align*} \binom{1000}{500} = \frac{1000!}{500!500!} \approx \frac{\sqrt{2\pi 1000}(1000/e)^{1000}}{(\sqrt{2\pi 500} (500/e)^{500})^2} = \frac{\sqrt{2\pi 1000}(1000/e)^{1000}}{2\pi 500 (500/e)^{1000}} \end{align*} (using sterling's approximation). Canceling out some terms will yield \begin{align*} &= \frac{\sqrt{1000}\cdot2^{1000}\cdot500^{1000}}{\sqrt{2\pi} 500 (500)^{1000}} = \frac{\sqrt{1000}\cdot2^{1000}}{\sqrt{2\pi} 500} = \frac{\sqrt{1000}\cdot4^{500}}{\sqrt{2\pi} 500} \end{align*} From here you can esimate values, i.e. $\sqrt{1000} \approx 31$, $\sqrt{2\pi} \approx \sqrt{6} \approx 2.5$, etc.

However, comparing orders of magnitude, the coefficient in front of $4^{500}$ hardly makes a difference, so we could conclude that $\binom{1000}{500}$ is by far closest to $4^{500}$ of the three choices given.