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Bonjour,

The problem i have follows from the definition of the binomial coefficient:

$\frac{n(n-1)...(n-k+1)}{k!} = {n \choose k}$

For 0$\leq{i}$ and i less than k, we observe that:

$\frac{n-i}{k-i}\geq\frac{n}{k}$

Is there a simple and intuitive arithmetic proof of this inequality ?

This inequality is sometimes used to prove that:

$(\frac{n}{k})^k\leq {n \choose k}$

  • 0
    You also need $k\leq n$.2012-11-30

2 Answers 2