Consider $s = \Theta(n^{\delta})$ for a $\delta\in (0,1)$ and let $p\in (0,1)$ with $m = \lfloor pn\rfloor$. Consider the random variable $Y$ which chooses $m$ elements from $\{1,\ldots,n\}$ such that any set of $m$ elements is equally likely. Then define $X$ to be $|Y \cap \{1,\ldots,s\}|$ where $|A|$ denotes the size of $A$. What is the total variation distance between $X$ and $\text{Bin}(s,p)$ in terms of the parameters $s,n,p$? (This is surely in the literature and using Stirling is tricky to work for all parameter ranges)



Coupling between two distributions
According to [formula (2)](https://arxiv.org/pdf/1905.03009), the total variation distance in question is bounded from above by $\dfrac{m-1}{n-1}$ assuming that $p=m/n$. Obviously, this bound does not depend on $s$. According to [this paper](ftp://ftp.stat.math.ethz.ch/U/hkuensch/hypergeom.pdf), this bound is optimal, up to a universal constant factor. 


(If you insist on having a bound for all values of $p\in[\frac mn,\frac{m+1}n)$, then I think it should be easy to additionally bound the total variation distance between two binomial distributions with the same value of $n$ and two different but close values of $p$, using the fact that the binomial family is stochastically monotone.)


