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I am trying to prove following inequality:

$$\binom{n}{k}<(en/k)^k$$

I tried Stirling approximation but I could not get anything further. Then I get $$\binom{n}{k}\approx \frac{\sqrt{2\pi n}n^n}{2\pi \sqrt{k(n-k)}(n-k)^{n-k}k^k}$$

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What have you tried? Please show your work so far. When you put in Stirling, you get some cancellation and a well known expression that goes to $e$. –  Ross Millikan Jun 2 '11 at 14:36
    
You may even be able to get the much sharper $\tbinom{n}{k}<e(n/k)^k$ –  Ross Millikan Jun 2 '11 at 14:49
2  
@Ross: For $n=10$ and $k=3$, ${n \choose k}=120$ but $e(n/k)^k < 101$. –  Shai Covo Jun 2 '11 at 15:32

1 Answer 1

$$\binom{n}{k} \left( \frac{k}{en} \right)^k = \frac{n(n-1) \ldots (n-k+1)}{n^k} \frac{k^k}{k! e^k} \leq \frac{k^k}{k! e^k} \text{ and since } e^k = \sum_m \frac{k^m}{m!},\;\;\; \frac{k^k}{k! e^k} < 1.$$

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I just edited your answer to enlarge it (it was so tiny!)...didn't change any notation or anything. I hope you don't mind? –  amWhy Jun 2 '11 at 21:34
    
No problem, thanks! –  Plop Jun 2 '11 at 21:40

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