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How many functions can be defined as A->B->C? I understand that there are |B|^|A| functions defined as A->B, and |C|^|B| functions defined as B->C, but for A->B->C, will there be |B|^|A| * |C|^|B| or (|C|^|B|)^(|B|^|A|) functions? Thank you!

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    Uh... so a function from $A$ to $B$ to $C$ is itself a function from $A$ to $C$ so there are clearly at most $|C|^{|A|}$ functions. This is obfuscated however based on the size of $|B|$. In particular if $|B|<|A|$ there will be fewer than the maximum number of functions. I don't have an answer off the top of my head but someone else may have experience with this problem that can chime in, else I'll think about it for a little while. – 2017-02-19
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    Do you mean "haw many functions from $A$ to $C^B$ ?" (where $C^B$ denotes the set of all functions from $B$ to $C$). – 2017-02-19
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    @Arden So it's something like function composition I guess, where f: A->B, g:B->C, and I want to know how many functions can be defined as g o f. Personally I think it's different from functions from A to C^B. But I could be wrong. If the case you are talking about is right, then the answer will be (C^B)^A, right? – 2017-02-19
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    @JMoravitz May I ask why will there be at most |C|^|A| functions? I'm thinking that if |A| = |C| = 1, and |B| =3, then somehow we could have more than 1 function defined as A->B->C. Hmm, is it right though? Thanks! – 2017-02-19
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    If you care about what path each element travels along, then yes there can be more. I interpret it however as only the start and end locations matter. In your example there is only the function where the element of A gets mapped to the element of C. – 2017-02-19
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    It sounds like you're asking: "How many functions $h : A \to C$ are there, which can be written as $h = g \circ f$, where $f : A \to B$ and $g : B \to C$?" Is that correct? – 2017-02-20

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Keeping in mind that the only thing that distinguishes one function as being "different" from another function is just how inputs get mapped to outputs. The "path" that an input travels along to get to its output does not matter in the slightest.

As such for $A=\{a_1,a_2\}, B=\{b_1,b_2\}, C=\{c_1\}$ there is only one function $\begin{cases} a_1\mapsto c_1\\ a_2\mapsto c_1\end{cases}$ despite there being many different "paths" that could arrive at this outcome.

The four functions $\begin{cases}a_1\mapsto b_1\mapsto c_1\\ a_2\mapsto b_1\mapsto c_1\end{cases},\begin{cases}a_1\mapsto b_1\mapsto c_1\\ a_2\mapsto b_2\mapsto c_1\end{cases},\begin{cases}a_1\mapsto b_2\mapsto c_1\\ a_2\mapsto b_1\mapsto c_1\end{cases},\begin{cases}a_1\mapsto b_2\mapsto c_1\\ a_2\mapsto b_2\mapsto c_1\end{cases}$ are all technically the same function in the end.

We are limited in the number of possible elements in the codomain being mapped to by the sizes of $|B|$ and $|A|$. There cannot be more elements in the image than either of $|B|$ or $|A|$ since $|Dom(f)|\geq |Rng(f)|$

Approaching via cases:

Suppose there are exactly $k$ elements in the image. Each of these elements has a set of elements as a preimage in $A$.

  • Find the number of unordered partitions of $A$ into exactly $k$ non-empty sets. This can be done in $\left\{\begin{smallmatrix}|A|\\k\end{smallmatrix}\right\}$ ways where $\left\{\begin{smallmatrix}n\\r\end{smallmatrix}\right\}$ is the stirling number of the second kind.

  • Pick which $k$ elements of $C$ are used in the range. This can be done in $\binom{|C|}{k}$ number of ways.

  • match each part in the partition to an element of the image via a permutation. This can be done in $k!$ number of ways.

Applying multiplication principle, there are exactly $\left\{\begin{smallmatrix}|A|\\k\end{smallmatrix}\right\}\binom{|C|}{k}k!$ functions from $A$ to $C$ with $k$ elements in the image.

Ranging over all possible values of $k$ and remembering that $k$ cannot exceed $|B|$ we have for a final total:

$$\sum\limits_{k=1}^{|B|}\left\{\begin{smallmatrix}|A|\\k\end{smallmatrix}\right\}\binom{|C|}{k}k!$$

(note: we could rewrite the top limit on the summation as $\min\{|A|,|B|,|C|\}$ but this is unnecessary because any $k$ which exceeds $|A|$ or $|C|$ will cause the term to be zero anyways in the calculation of the stirling number or the binomial coefficient)