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?
${2n-l\choose n-k}\leq {2n-l \choose \frac{2n-l}{2}}$
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?
${2n-l\choose n-k}\leq {2n-l \choose \frac{2n-l}{2}}$
HINT
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 $\begin{cases} \binom{2N}{r} < \binom{2N}{r+1} & \text{when } r \leq N-1\\ \binom{2N}{r+1} < \binom{2N}{r} & \text{when } r \geq N \end{cases}$
Hint: Calculate
$\frac{{2n-l\choose m+1}}{{2n-l\choose m}}$
Can you find out when the fraction is more/less than 1?