I would like to ask, if my intuitive understanding of how to count is correct. Please elaborate or correct me, if the below interpretation is not sound.
Should constructing a $r$-sample (unordered set) without replacement from a population of size $n$ be equivalent to allocating $r$ indist. balls to $n$ distinguishable cells? Atleast, that's how I picture it even visually.
Consider the random placement of $r$ balls in $n$ distinguishable cells.
$$ \begin{array}{c|lcr} & \text{No restriction} & \text{Cell can be at most singly-occupied} \\ \hline r \text{ dist. balls} & n^r & (n)_r = n(n-1)\cdots(n-r+1)=\frac{n!}{r!}\\ r \text{ indist. balls} & {n + r - 1 \choose r} & {n \choose r} \end{array} $$