3Finding minimum $f(k)$ ($n$: fixed natural number and $k=0,1,2,\cdots,n-1$ I would appreciate if somebody could help me with the following problem
Q. Finding minimum $f(k)$ where $n \in \mathbb N$ and $k = 0,1,...,n-1$.
$$f(k)={2k+1 \choose k} \times {2n-2k-1\choose n-k}$$
I tried and find 
$n=6$ then $f(2),f(3)$ has minimum 
$n=7$ then $f(3)$ has minimum 
 A: It’s convenient to let $m=n-1$ and define
$$g_m(k)=\binom{2k+1}k\binom{2m+1-2k}{m+1-k}$$
for $k=0,\ldots,m$; we wish to choose $k$ so as to minimize $g_m(k)$.
Let $\ell=m-k$; then
$$g_m(k)=\binom{2k+1}k\binom{2m+1-2k}{m+1-k}=\binom{2m+1-2\ell}{m+1-\ell}\binom{2\ell+1}\ell=g_m(\ell)=g_m(m-k)\;.$$
That is, $g_m$ is symmetric over the range $k=0,\ldots,m$. Now
$$\begin{align*}
g_m(k+1)-g_m(k)&=\binom{2k+3}{k+1}\binom{2m-1-2k}{m-k}-\binom{2k+1}k\binom{2m+1-2k}{m+1-k}\\\\
&=\binom{2k+3}{k+1}\binom{2m-1-2k}{m-k}-\binom{2k+1}{k+1}\binom{2m+1-2k}{m-k}\\\\
&=\left(\frac{(2k+3)(2k+2)}{(k+2)(k+1)}-\frac{(2m-2k)(2m+1-2k)}{(m+1-k)(m-k)}\right)\binom{2k+1}{k+1}\binom{2m-1-2k}{m-k}\\\\
&=2\left(\frac{2k+3}{k+2}-\frac{2m+1-2k}{m+1-k}\right)\binom{2k+1}{k+1}\binom{2m-1-2k}{m-k}\\\\
&=\frac{2(2k+1-m)}{(k+2)(m+1-k)}\binom{2k+1}{k+1}\binom{2m-1-2k}{m-k}\;,
\end{align*}$$
which has the same algebraic sign as $2k+1-m$. 
In particular, $g_m(k)>g_m(k+1)$ if $2k+1<m$, $g_m(k)=g_m(k+1)$ if $2k+1=m$, and $g_m(k)<g_m(k+1)$ if $2k+1>m$. Thus, $m=2r+1$, then $g_m(k)$ attains its minimum at $k=r$ and $k=r+1$, and if $m=2r$, then $g_m(k)$ attains its minimum at $k=r$.
Alternatively, $g_m$ attains its minimum at $k=\left\lfloor\frac{m}2\right\rfloor$ and at $k=\left\lceil\frac{m}2\right\rceil$.
