We know that stirling number second kind, which has the recurrence relation as $f(n, k) = f(n - 1, k - 1) + k \times f(n - 1, k)$, counts the number of ways to put n elements in k non-empty partition.
If every partition contains at least r elements, what would be the recurrence? I'm guessing that would be $$f(n, k) = \binom{n - 1}{r - 1} \times f(n - r, k - 1) + k \times f(n - r, k)$$
Is that right? If there's any other way, please mention that too.
Thanks in advance!
The second part should be $kf(n-1,k)$ because you are taking one element and placing it in the current $k$ partitions.
This numbers are called Associated Stirling numbers of the second kind.
As an introduction point see https://oeis.org/A008299 when $r=2.$