I need to find $$ \max_q\binom{n-(q-1)d}{q} $$

I understand $\binom{n}{k}$ reaches its maximum when $k=\lfloor\frac{n}{2}\rfloor$. But I don't think this holds here since if $q$ gets bigger then $n-(q-1)d$ gets lower.

edit When $d$ is a fixed positive integer and $q$ is an integer such that $0\leq q \leq \frac{n}{d+1}$

  • $\begingroup$ You should specify who $q$ and $d$ are. $\endgroup$ – Giuseppe Negro Mar 14 '18 at 8:57

Stirling's approximation, zeroth order, gives $$\binom{n-(q-1)d}{q} = \frac{(n-(q-1)d)!}{q!(n-(q-1)d-q)!} \\ \approx \frac{\sqrt{(n-(q-1)d)} (\frac{(n-(q-1)d)}{e})^{(n-(q-1)d)}}{\sqrt{2\pi q(n-(q-1)d-q)} (\frac{q}{e})^q (\frac{(n-(q-1)d-q)}{e})^{(n-(q-1)d-q)}} \\ = \frac{1}{\sqrt{2\pi}} \frac{(n-(q-1)d)^{n-(q-1)d+\frac12}}{q^{q+\frac12} (n-(q-1)d-q)^{n-(q-1)d-q+\frac12}} e^{(q-1)(n-(q-1)d)-q^2} \\ $$

Now differentiate with respect to $q$...


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