From easy computation we can get
\begin{align*}
{n\choose{\ell}}=\frac{n}{\ell}\cdot\frac{n-1}{\ell-1}\cdots\frac{n-(\ell-1)}{1}\geq\frac{n^{\ell}}{\ell^{\ell}},
\end{align*}
where the last inequality follows from the fact that each of these
$\ell$ terms in the product is at least $\frac{n}{\ell}$ and $n\geq \ell>0.$
So, I just wonder if we could do better than this lower bound $\displaystyle\frac{n^{\ell}}{\ell^{\ell}}$ for the binomial coefficient.
Any comments or advice would be appreciated. Thanks for patient reading.
Best Answer
I do not remember the reference of the paper,; from my notes :
Using $$H(x)=-x\log(x)-(1-x)\log (1-x)$$ very tight bounds
$$\text{lower}=\sqrt{\frac{n}{8 l (n-l)}}\,\exp\Bigg[n\, H\left(\frac{l}{n}\right) \Bigg]\leq \binom{n}{l}\leq \sqrt{\frac{n}{2\pi l (n-l)}}\,\exp\Bigg[n\, H\left(\frac{l}{n}\right) \Bigg]=\text{upper}$$
Trying for $n=20$ $$\left( \begin{array}{cccc} l & \text{lower} & \binom{n}{l}& \text{upper} \\ 1 & 19 & 20 & 22 \\ 2 & 176 & 190 & 198 \\ 3 & 1039 & 1140 & 1173 \\ 4 & 4389 & 4845 & 4952 \\ 5 & 13990 & 15504 & 15786 \\ 6 & 34892 & 38760 & 39372 \\ 7 & 69679 & 77520 & 78624 \\ 8 & 113120 & 125970 & 127642 \\ 9 & 150748 & 167960 & 170100 \\ 10 & 165794 & 184756 & 187079 \end{array} \right)$$
This paper did improve.