Maximizing $\left.\binom{N-40}{60-13}\middle/{\binom{N}{60}}\right.$

Question

I want to choose $N \in \mathbb{N}\geq 87$ to maximize $$\frac{\dbinom{N-40}{60-13}}{\dbinom{N}{60}} = \frac{(N-40)!60!(N-60)!}{47!(N-87)!N!}$$ which is equivalent to maximizing $$\frac{(N-40)!(N-60)!}{(N-87)!N!}= \frac{(N-60)\cdots(N-86)}{N\cdots(N-39)}$$

Clearly the denominator "grows faster" than the numerator, so intuitively we want to make $N$ as small as possible, ie $N = 87$. Is there some way to show this rigorously, without taking derivatives?

Answer 1

Let $f(N) = \frac{(N-60)\cdots(N-86)}{N\cdots(N-39)}$

$F(N+1) = \frac{(N-59)\cdots(N-85)}{(N+1)\cdots(N-40)}= \frac{N-59}{N+1}\frac{(N-60)\cdots(N-86)}{N\cdots(N-39)}\frac{N-40}{N-86}=\frac{N-59}{N+1}\frac{N-40}{N-86}F(N)$

As $N\ge87$ we just have to show $\frac{N-59}{N+1}\frac{N-40}{N-86}<1$

Which happens iff $(N-59)(N-40) < (N+1)(N- 86)$ iff $N \ge 175$.

So $F(N)$ is increasing from $86 \le N \le 174$ and decreasing on $175 \le N$. So max will be at the last point it is increasing. i.e. at $N = 174$.