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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:

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

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You also need $k\leq n$. –  Alex R. Nov 30 '12 at 17:52
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2 Answers

This is pretty straight forward. Note that $\frac{n-i}{k-i}\geq \frac{n}{k}$ iff $nk-ik\geq nk-ni$, which is true iff $-ik\geq -in$ which again is true (for our $i$) iff $k\leq n$, and this gives the answer.

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user45150: thank you very much ! –  user43414 Nov 30 '12 at 17:57
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If $0 < i < k < n$, then $$\frac{n-i}{k-i} = \frac{n}{k}\times\frac{1 - \frac{i}{n}}{1-\frac{i}{k}} > \frac{n}{k}$$ since $1-\frac{i}{n} < 1 - \frac{i}{k}$. For an intuitive explanation let us look at a set of $n$ balls of which $k$ balls are blue and $n-k$ are not blue. Then, the proportion of blue balls in the set is $\frac{k}{n}$. Throw away $i$ blue balls. Then, the proportion of blue balls in the reduced set is smaller because you have thrown away proportionately far more blue balls than non-blue balls. That is, $$\frac{k-i}{n-i} < \frac{k}{n} \Rightarrow \frac{n}{k} < \frac{n-i}{k-i}.$$

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thank you. To resort to set-theory is always useful. –  user43414 Nov 30 '12 at 20:23
    
I'll take a close look to this ! –  user43414 Nov 30 '12 at 20:24
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