Find the condition where the product of two factorials is minimized

Question

I have to find the maximum value of a combination ${N \choose H} = \dfrac{N!}{H!(N-H)!}$ where $1 \leq H \leq N \leq 16$. My reasoning was to find the maximum value you have to maximize the numerator $N!$ and minimize the denominator $H!(N-H)!$. The maximum value of numerator occurs when $N = 16$. Then the term becomes, $\dfrac{16!}{H!(16-H)!}$. As, $1 \leq H \leq 16$, I can manually check for which $H$ value the term is maximized. But is there any other way to find out that the denominator is minimized when $H = 8$?

I tried to break now into three conditions, namely when $H > (16-H)$, $H < (16 - H)$ and $ H = 16 - H$, but I couldn't establish the relation between the three conditions. Any help would be appreaciated.

Answer 1

From the complementary combination $\displaystyle\binom{N}{H} = \binom{N}{N-H}$ we conclude that

$\displaystyle\binom{N}{H}$ is symmetric with respect to $\displaystyle H=\bigg\lceil{\frac{N}{2}}\bigg\rceil$

Now we show that $\displaystyle\binom{N}{H}$ is monotonically increasing for $\displaystyle 0\le H \le \bigg\lceil{\frac{N}{2}}\bigg\rceil$ keeping $N$ fixed. Equivalently we will show $$\text{If } \displaystyle 0\le H_1 < H_2\le \bigg\lceil{\frac{N}{2}}\bigg\rceil \text{ then } \binom{N}{H_1} <\binom{N}{H_2}$$

Proof: Let $H_2 = H_1 + k$ with $k\ge1$. Now $$\begin{equation*} \begin{aligned} \binom{N}{H_2} &= \frac{N\cdot(N-1)\cdots(N-H_2+1)}{1\cdot2\cdot3\cdots H_2}\\ &=\frac{N\cdot(N-1)\cdots (N-H_1+1)\cdot(N-H_1)\cdot(N-H_1-1)\cdots(N-H_1-k+1)}{1\cdot 2\cdot 3\cdots H_1\cdot(H_1+1)\cdots(H_1+k)} \\ &=\binom{N}{H_1}\cdot\frac{(N-H_1)(N-H_1-1)\cdots(N-H_1-k+1)}{(H_1+1)(H_1+2)\cdots(H_1+k)}\\ &=\binom{N}{H_1}\cdot\frac{N-H_1}{H_1+k}\cdot\frac{N-H_1-1}{H_1+k-1}\cdots\frac{N-H_1-k+1}{H_1+1} \end{aligned} \end{equation*}$$ We need to show now every fraction of RHS $\ge1$ and at least one fraction $>1$

Notice that, every fraction of RHS is of the form $\displaystyle \frac{N-H_1-r}{H_1+k-r}$ where $0\le r \le k-1$

$$\begin{equation}\displaystyle \begin{aligned} \frac{N-H_1-r}{H_1+k-r}&=\frac{N-H_2+k-r}{H_2-r}\\ &\ge \frac{N-H_2+1}{H_2-r} \text{ as } k-r\ge1 \\ &\ge \frac{N-\bigg\lceil\displaystyle\frac{N}{2}\bigg\rceil+1}{\bigg\lceil\displaystyle\frac{N}{2}\bigg\rceil-r} \\ &=\frac{\bigg\lfloor\displaystyle\frac{N}{2}\bigg\rfloor+1}{\bigg\lceil\displaystyle\frac{N}{2}\bigg\rceil-r} \end{aligned} \end{equation}$$ If $r > 0$ this fraction $>1$ and if $r=0$ this fraction $\ge1$ and we are done. $\blacksquare$

Answer 2

Hint:

Consider the ratio: $$ \frac{\binom N{H+1}}{\binom NH}$$ and compare it with 1.