maximum of a product of binomial coefficients

Let $$x,b,\ell$$ non negative integers, with $$\ell\le b.$$ Consider $$b,\ell$$ fixed. Let also $$F_b(x) = \binom{b}{x}\binom{b}{\ell-x}.$$ The maximum of $$F_b(x)$$ for $$x=0,1,2,\dots, \ell$$ is $$F_b(\lfloor\ell/2\rfloor).$$ A proof of this fact can be done in two steps.If $$G_b(x)=\frac{F_b(x)}{F_b(x+1)}$$ for $$0\leq x\leq \lfloor \ell/2\rfloor-1$$ then we can prove that $$G_b(x)$$ is increasing so $$G_b(x)\le G_b(\lfloor \ell/2\rfloor-1).$$ Also we can prove that $$G_b(\lfloor \ell/2\rfloor-1)\leq 1.$$ Now follows that $$\max F_b(x) =F_b(\lfloor \ell/2\rfloor).$$ The previous are elementary but a bit technical. Although, I believe there is a simpler proof and not so technical. Any idea?

• Just an idea: your function is related to the probability mass function of the hypergeometric distribution provided that you set $K=b$, $N=2K=2b$, $k=x$, $n=l$ there. You could check also if this gives you some idea. – BillyJoe Jan 12 at 16:44

Your function is related to the probability mass function of the hypergeometric distribution provided that you set in that page:

• $$N=2b$$ the population size,
• $$K=b$$ the number of success states in the population,
• $$n=l$$ the number of draws (i.e. quantity drawn in each trial),
• $$k=x$$ the number of observed successes.

Under the section "Combinatorial identities" of the cited page you find the following identity:

$$\frac{{K \choose k}{N-k \choose n-k}}{{N \choose n}}=\frac{{n \choose k}{N-n \choose K-k}}{{N \choose K}}$$

$$\frac{{b \choose x}{2b-b \choose l-x}}{{2b \choose l}}=\frac{{l \choose x}{2b-l \choose b-x}}{{2b \choose b}}$$
Now we know that both factors of the numerator of the RHS are maximized with $$x = \lfloor l/2 \rfloor$$ or $$x = \lceil l/2 \rceil$$.