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?
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1Just 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. – Fabius Wiesner Jan 12 '20 at 16:44
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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:
$$\frac{{K \choose k}{N-k \choose n-k}}{{N \choose n}}=\frac{{n \choose k}{N-n \choose K-k}}{{N \choose K}}$$
which in your case is:
$$\frac{{b \choose x}{2b-b \choose l-x}}{{2b \choose l}}=\frac{{l \choose x}{2b-l \choose b-x}}{{2b \choose b}}$$
Now we know that both factors of the numerator of the RHS are maximized with $x = \lfloor l/2 \rfloor$ or $x = \lceil l/2 \rceil$.
Fabius Wiesner
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Yes. In fact you only need the "combinatorial" identity, without referring to hypergeometric distribution. Thanks. – 111 Jan 12 '20 at 23:04