asymptotic probability of at least $m$ unique numbers in selection of $k$ integers in range $1, \ldots, n$ The starting point is the following:
Given the positive fractions $0 \lt c \le d \le 1$, what is the probability of getting at least $m=cn$ unique numbers when selecting $k=dn$ integers independently and uniformly from the range $[1,n]$?
In a related question, an exact answer for a fixed $m$ is given as a sum involving binomial coefficients:
https://math.stackexchange.com/a/1087968/471924
Instead of an exact answer, I would like to consider where $m$ is significantly smaller than the expectation value, and to discuss the asymptotics of the probability as $n$ increases without bound. 
In particular I am looking for conditions on $c$ and $d$ that allow a statement like:
With probability $1-o(1)$ there are at least $cn$ unique numbers when randomly selecting $dn$ integers uniformly from the range $[1,n]$.
Is there someway to get from the exact results to conditions on $c,d$ which would make the above asymptotic statement true?
 A: The number $Z_{dn} $ of unique numbers when selecting $dn$ numbers with replacement from $n$ is about $(1-e^{-d})n$ on average.  Moreover, the variance is of order $n$, so $Z_{dn} = (1-e^{-d})n + O_p(\sqrt{n})$. 
It can be shown that the error of this approximation is asymptotically normal. Here is the idea. Let the numbers be drawn one by one. Denote by $T_k$ the time of $k$th unique number appearing. Then $T_k = G_1 + \dots + G_k$, where $G_k$ are independent with $G_k\simeq \mathrm{Geo}(\frac{n-k+1}{n})$. Despite they have different distribution, the CLT (e.g. in the Lindeberg form) can be applied for $k = cn$, $c\in(0,1)$. In order to deduce the asymptotic normality for $Z$ from here, notice that $\{Z_{m}\ge k\} = \{T_k\le m\}$.
TLDR: you have the desired statement for any $c<1-e^{-d}$. 

Let $A_i = \{\text{the number $i$ is chosen}\}$. Then the expectation of $Z_{nd}$ is
$$
\mathrm{E}[Z_{nd}] = \mathrm{E}\Big[ \sum_{i=1}^{n} \mathbf{1}_{A_i}\Big] = \sum_{i=1}^n \left(1 - \mathrm{P}(A_i^c)\right) = n \left(1 - \Big(1-\frac1n\Big)^{nd}\right)\approx n (1-e^{-d}).
$$
