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Let $m \leq n, n \leq N$ and $0\leq k \leq m$.

I am wondering what is the dependence of $n$ and $N$ that for all $m, k$ $$ \frac{{N-m \choose n-k}}{{N \choose n}}\leq 1. $$

Thank you for your help.

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Do you want that to be true for all $m$ and $k$ meeting the conditions? –  Henry Nov 29 '12 at 21:54
    
Thank you. Yes, for all $m,k$. –  Michael Nov 29 '12 at 22:30

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If you try and force the numerator to be greater than the denominator which means making $N-m$ large relative to $n-k$, this happens at $m=k$. Here we have $$\frac{\binom{N-k}{n-k}}{\binom{N}{n}} = \frac{\binom{N-k}{n}}{\binom{N}{n}}\;.$$ This is clearly $ < 1$ and so for fixed $N,n$ this inequality is true for all $k,m$ given your criteria. In fact the inequality can be improved to $ <1 $ as mentioned above.

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This is wrong. My apologies. $\binom{n}{k}$ is maximized at $k = \lfloor \frac{n}{2} \rfloor$ and $ \lceil \frac{n}{2} \rceil$. –  Patrick Nov 29 '12 at 23:24
    
Thank you, I am curious now, what would be the solution if $k$ and $m$ are fixed. What the relationship between $N$ and $n$ should be to satisfy inequality? Thank you. –  Michael Nov 30 '12 at 15:47

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