Approximation of $\frac {n-c \choose k} {n\choose k} $ using a radical

Question

Is there a good approximation for $\frac {n-c \choose k} {n\choose k} $, given large parameter n (k,c can be large as well, but k,c<n/2)? I think I can express it as a simple exponent function f(n,k,c), but i am not sure.

I prove that this expression equals $\Pi_{j=0}^{k-1}(1-\frac{c}{n-j})$ But i can't continue further

Answer 1

Using the factorial definition, we can rewrite this as $$\frac{n!(n-(c+k))!}{(n-c)!(n-k)!}$$

and then use falling factorials to reduce to $$\frac{n^{\underline{(c+k)}}(n-(c+k))!^2}{(n-k)^\underline c (n-c)^\underline k (n-(c+k))!^2}=\frac{n^{\underline{c+k}}}{(n-k)^\underline c (n-c)^\underline k}$$

$n^\underline t$ is a polynomial of degree $t$. Hence, we may apply what I refer to as Domination Leads To Irrelevancy: $$\lim_{x\to\infty}\bigg[\frac{\sum_{k=0}^m a_k x^k}{\sum_{k=0}^n b_k x^k}\bigg]=\lim_{x\to\infty}\bigg[\frac{a_m}{b_n}x^{m-n}\bigg]=\begin{cases} \frac{a_mb_n}{|a_mb_n|}\infty & m>n \\ \frac{a_m}{b_n} & m=n \\ 0 & m<n \end{cases}$$

to see that the value will be very close to, and approaches $1$ for sufficiently large $n$ as both polynomials are degree $c+k$ and our leading coefficients are both $1$.

Answer 2

You can get a simple upper bound by taking $$ \binom{n}{k} < \frac{n^k}{k!}\\ \binom{n-c}{k} > (n-c-k+1)^{k}{k!} $$