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The question is, how many functions are there from $A$ to $\mathcal{P}(B)$ such that $$\bigcup_{x\in A} f(x) = B.$$ Both sets are finite. The only way I can come up with is to count by hand (small sets in my case) but I was wondering if this is a known problem with a known solution.

Another way to pose the problem is: Count the number of coverings of the set $B$ with exactly $N$ labeled subsets (where $N$ is the size of $A$), but I don't see that helping because then one has to start counting permutations and cases where two subsets are repeated, etc.

It seems related to the set covering problem and maximum coverage problem, and I have read this question which is also related, but I thought maybe my version is a bit easier/more difficult?

I just came up with a possibility for a recursion: $n_{A,B} = |A|^{2^{|B|}} - \sum_{S\subset B} n_{A,B/S}$. Does that sound right? Final edit: No it doesn't... I'm still repeating subsets everywhere.

Thank you very much for your help!

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Take $y\in B$ and $C_y=\{x\in A:y\in f(x)\}$, which is a subset of $A$

Then notice that the only undesirable case is $C_y=\emptyset$, only in that case you will have $y\notin \bigcup_{x\in A}f(x)$. So you can chose $C_y$ in $2^{|A|}-1$ ways.

Repeat this for all $y\in B$ and you will get your number is

$$\left(2^{|A|}-1\right)^{|B|}$$

EDIT: To explain it in words, each element of $B$ has to be covered by $f$ of each element of a non-empty subset of elements of $A$, and the choice of these non-empty subsets uniquely defines $f$

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    I think there is a problem with this solution: This lets me pick all $C_y$ the same (for example a singleton in $A$) and then I would not have a function defined in all elements of $A$. Am I correct in saying this? – 2017-01-14
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    There is no problem in having $f(x)=B$, and $f(x')=\emptyset$ for $x'\ne x$. This is a solution too. $\bigcup_{x\in A}f(x)=B$ – 2017-01-14
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    You are right, it's just my wording was incorrect, $f(x)$ can never be the empty set... Sorry for this, it comes from the interpretation of a text and it's hard to translate precisely. But this gave me a head start and it answers the question so I will accept it! Thank you! – 2017-01-14