Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I today see a approximated equation, when $n \ll u $:

$$\log {u \choose n} \approx n \Big(\log \frac{u}{n} + 1.44\Big)$$

I would like to know how to prove it.

share|cite|improve this question… – pedja Apr 24 '12 at 8:47
Use Stirling's approximation for $n!$. – Johannes Kloos Apr 24 '12 at 9:05
@JohnBentin No, I mean n is smaller than u. – Fan Zhang Apr 24 '12 at 16:12

1 Answer 1

up vote 0 down vote accepted

It looks like pedja has pointed you in the right direction here. On Wikipedia there's the bounds: \begin{align*} {u \choose n} &= \frac{1}{n!} u(u-1)\cdots(u-n+1) \\ & \leq \frac{u^n}{n!} \\ & \leq \left(\frac{u\ e}{n}\right)^n \end{align*} since $n! \geq \left(\frac{n}{e}\right)^n$, and \begin{align*} {u \choose n} & = \frac{u}{n} \cdot \frac{u-1}{n-1} \cdots \frac{u-n+1}{1} \\ & \geq \left(\frac{u}{n}\right)^n. \end{align*}

By taking $\log_2$ of both sides of these bounds, we obtain \[n \log_2 \frac{u}{n} \leq \log_2 {u \choose n} \leq n \left(\log_2 \frac{u}{n}+\log_2 e\right),\] where $\log_2 e$ is approximately $1.44$.

If $n=o(\log u)$, then \[\log_2 {u \choose n} \sim n \log_2 \frac{u}{n},\] so the $1.44$ won't be necessary.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.