Number of functions $\{ f: [k] \rightarrow[n] :|f([k])|=m \}$

Asked Jan 10 '21 at 14:35

Active Jan 10 '21 at 16:08

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If we have $k,\;n,\;m\in\mathbb{N}: m\leq n$, how to determine $|\{ f: [k] \rightarrow[n] :|f([k])|=m \}|$. Were we have $ [n]:=\{1,2,...,n\}.$

So firstly, there should be a $ \binom{n}{m} $ ways to choose a subset of $[n]$ with the cardinality $m$.

But this is where I am stuck, I would appreciate any help for further calcualtion.

edited Jan 10 '21 at 16:08

N. F. Taussig

asked Jan 10 '21 at 14:35

variableXYZ

1

The relevant notion here is Sterling numbers of the second kind – lulu Jan 10 '21 at 14:52
Stirling actually – ancient mathematician Jan 10 '21 at 14:52
@ancientmathematician I am aware of that, but not sure how to calculate it. – variableXYZ Jan 10 '21 at 15:34
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the wikipedia page is a good place to start https://en.m.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind – kodlu Jan 10 '21 at 20:01
You can also use an inclusion-exclusion argument; the last part of this answer does it for the case $m=n$, and you already know how to modify that result. – Brian M. Scott Jan 10 '21 at 21:30

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