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Let $n, l \in \mathbb{N}, l\leq n$ be fixed. Let $k\in \mathbb{N}$ with $0 \leq k \leq l$. How to show the following? $$ {2n-l\choose n-k}\leq {2n-l \choose \frac{2n-l}{2}} $$

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Is $l$ even?${}$ – wj32 Nov 4 '12 at 4:29
No, $l$ is any. – user202312 Nov 4 '12 at 4:29
So you're using the Gamma function definition of $\binom{7}{7/2}$? – wj32 Nov 4 '12 at 4:31
Yes, or just a Stirling formula. – user202312 Nov 4 '12 at 4:35
up vote 0 down vote accepted


Prove that $$\displaystyle \begin{cases} \binom{2N}{r} < \binom{2N}{r+1} & \text{when } r \leq N-1\\ \binom{2N}{r+1} < \binom{2N}{r} & \text{when } r \geq N \end{cases}$$

We have that $$\binom{2N}{r+1} = \binom{2N}{r} \times \left( \frac{2N-r}{r+1} \right).$$ Note that $$ \begin{cases} \left( \frac{2N -r}{r+1} \right) > 1 & r \leq N - 1\\ \left( \frac{2N -r}{r+1} \right) < 1 & r \geq N \end{cases}$$ Hence, we have that $$\begin{cases} \binom{2N}{r} < \binom{2N}{r+1} & \text{when } r \leq N-1\\ \binom{2N}{r+1} < \binom{2N}{r} & \text{when } r \geq N \end{cases}$$

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Tank you, but I stil don't see how to use it for my inequality. Could you elaborate, please. Thank you. – user202312 Nov 4 '12 at 4:43
@user202312 In your case, choose $N = 2n-l$. Essentially, you want to prove that the middle binomial coefficient is the largest. – user17762 Nov 4 '12 at 5:06

Hint: Calculate

$$\frac{{2n-l\choose m+1}}{{2n-l\choose m}}$$

Can you find out when the fraction is more/less than 1?

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