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  • $\begingroup$ Thank you @Max for your answer. It seems that in your solution $m$ is the number of unique selected elements, is it right? $\endgroup$ Commented May 10, 2020 at 13:27
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    $\begingroup$ @CarlosA.AstudilloTrujillo: Yes. Selection with replacement is similar - added it now $\endgroup$ Commented May 10, 2020 at 13:43
  • $\begingroup$ Another option to answer my initial question is just to consider $m$ as $m^{′}$ in your formula without replacement and calculate $m{′}=N{\cdot}k\left(1−B_{0}(n,p)\right)$, where $B_{r}(n,p)$ is the binomial distribution with parameters $n=m$ (as defined in the initial question), $p=\frac{1}{N{\cdot}k}$ and $r$ is the number of times an element in $S_k$ is selected. Thus, $1−B_{0}(n,p)$ is the probability that an element of $S_{k}$ is selected at least once in $n$ trials. $\endgroup$ Commented May 10, 2020 at 14:30
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    $\begingroup$ I tested your solution of selection with replacement and it is more accurate than my previous suggestion of considering $m^{'}$ and selection without replacement. Thank you again. $\endgroup$ Commented May 10, 2020 at 15:22