dividing $\binom{n-m-1}{k}$ by $\binom{n-1}{k}$

Question

for $n,m\in \mathbb{N}$ and $m<n$, is there a closed form for this division? $\frac{\binom{n-m-1}{k}}{ \binom{n-1}{k}}$.

I ended up with $\frac{(n-m-1)!}{(-nk+mk+k)!}\times \frac{(-nk+k)!}{(n-1)!}$ and couldn't simplify it more. The case where $m=1$ I remember I found it $1-\frac{k}{n-1}$.

It seems you might find useful to check the definition of the binomial coefficients since the formula you suggest is way offbase. — Did, Sep 27 '16 at 19:30
@Did do you mean this one $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ ? — seteropere, Sep 27 '16 at 19:32
@Did how about being specific instead of sending me to a blocked youtube in Canada. I applied that one and got to this point after removing $k!$ from both terms. — seteropere, Sep 27 '16 at 19:37
Sorry for the link, what about this one? Anyway, I WAS specific, thank you, and the mystery is how you reached these $(-nk+mk+k)!$ and $(-nk+k)!$... things. Care to explain? — Did, Sep 27 '16 at 19:41
It seems you misunderstood the formula: $k!(n-k)!\neq (kn-k^2)!$ — N74, Sep 27 '16 at 19:41
@N74 I miscalculated the term $(n-m-1-k)!$ thanks for being helpful anyways. — seteropere, Sep 27 '16 at 19:43

score 2 · Accepted Answer · answered Sep 27 '16 at 19:40

Your arithmetic after cancelling the factors of $k!$ seems to have gone astray.

$$\begin{align*} \frac{\binom{n-m-1}k}{\binom{n-1}k}&=\frac{(n-m-1)!k!(n-1-k)!}{k!(n-m-1-k)!(n-1)!}\\ &=\frac{(n-m-1)!(n-1-k)!}{(n-m-1-k)!(n-1)!}\\ &=\frac{(n-m-1)^{\underline k}}{(n-1)^{\underline k}}\\ &=\prod_{i=1}^k\frac{n-m-i}{n-i}\\ &=\prod_{i=1}^k\left(1-\frac{m}{n-i}\right) \end{align*}$$

Here $x^{\underline k}=x(x-1)(x-2)\ldots(x-k+1)$ is a falling factorial.

dividing $\binom{n-m-1}{k}$ by $\binom{n-1}{k}$

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