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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 $\binom{n}{k}\approx \frac{\sqrt{2\pi n}n^n}{2\pi \sqrt{k(n-k)}(n-k)^{n-k}k^k}$

  • 2
    @Ross: For $n=10$ and $k=3$, ${n \choose k}=120$ but $e(n/k)^k < 101$.2011-06-02

1 Answers 1

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