Get number of functions f from A to B where F(X) $\supseteq$ Y, Y $\subseteq$ B

Question

$X\subseteq A$ and $Y\subseteq B$ are finite sets. $|A| = m, |B| = n, |X| = k$, and $|Y| = \ell$. I want to get the number of functions from $A$ to $B$ where $F(X) \supseteq Y$. I don't know $|F(X)|$.

I think that if $F(X) \supseteq Y$ then $\ell$ must be less than or equal to $k$ because $|F(X)| \leq |X|$.

Can someone help me?

Hint: How many functions are there from a set with $u$ elements to a set with $v$ elements? — Hans Hüttel, Nov 10 '16 at 22:13
@HansHüttel There are $v^u$ functions. How can this help me? — Johny, Nov 11 '16 at 12:49
You know that $|X| = k$. So what are the possible cardinalities of $F(X)$?? — Hans Hüttel, Nov 11 '16 at 13:10
Yes. Now remember that $F(X) \supseteq Y$ and that a function must also assign values to the elements of $A \setminus X$. How many elements are there in this set? — Hans Hüttel, Nov 11 '16 at 13:18
In set $A \setminus X$ are elements $x$ that $x \in A$ and $x \notin X$, so $m - k$? — Johny, Nov 11 '16 at 13:26
@HansHüttel And $|Y|$ must be less than or equal to $|F(X)|$ because $F(X) \supseteq Y$. So $|F(X)| = <l; k>$. (Or not?) I still don't know how to solve it :( — Johny, Nov 11 '16 at 17:51

score 2 · Accepted Answer · answered Nov 13 '16 at 19:13

This gets rather messy. Say that a function $f:A\to B$ is good if it has the desired properties, and suppose that $f$ is a good function. Then $f$ maps $X$ onto $Y$, but it may also send some points of $X$ to points in $B\setminus Y$. For each $i$ let $\mathscr{F}_i$ be the set of good functions $f$ such that

$$\{x\in X:f(x)\in Y\}|=i\;;$$

clearly $i\ge\ell$, since $f$ has to map $X$ onto $Y$, and of course $i\le k=|A|$. I’ll count the functions in $\mathscr{F}_i$.

Suppose that $f\in\mathscr{F}_i$; if $a\in A\setminus X$, $f(a)$ can be any point of $B$, so there are $n^{m-k}$ ways to define $f\upharpoonright(A\setminus X)$, the restriction of $f$ to $A\setminus X$. There are $\binom{k}i$ ways to choose an $i$-element subset $S$ of $X$ to be mapped onto $Y$, and the $k-i$ elements of $X\setminus S$ can be mapped to $B\setminus Y$ in $(n-\ell)^{k-i}$ ways. There are ${i\brace\ell}$ ways to partition $S$ into $\ell$ non-empty parts, where ${i\brace\ell}$ is a Stirling number of the second kind, and there are $\ell!$ ways to assign these parts to be the preimages of the $\ell$ members of $Y$, so there are $\ell!{i\brace\ell}$ maps of $S$ onto $Y$. Thus,

$$|\mathscr{F}_i|=n^{m-k}(n-\ell)^{k-i}\binom{k}i\ell!{i\brace\ell}\;,$$

and there are therefore

$$\sum_{i=\ell}^k|\mathscr{F}_i|=\ell!\,n^{m-k}\sum_{i=\ell}^k\binom{k}i{i\brace\ell}(n-\ell)^{k-i}$$

good functions. I don’t at the moment see any nice simplification of this, but the following observations may be of some use. Let

$$g(x)=\sum_{j\ge 0}{j\brace\ell}\frac{x^j}{j!}=\frac{(e^x-1)^\ell}{\ell!}$$

and

$$h(x)=\sum_{j\ge 0}(n-\ell)^j\frac{x^j}{j!}=e^{(n-\ell)x}\;;$$

then the coefficient of $\frac{x^k}{k!}$ in

$$\frac{e^{(n-\ell)x}(e^x-1)^\ell}{\ell!}=\left(\sum_{j\ge 0}{j\brace\ell}\frac{x^j}{j!}\right)\left(\sum_{j\ge 0}\frac{x^j}{j!}\right)$$

is

$$\sum_{i=\ell}^k\binom{k}i{i\brace\ell}(n-\ell)^{k-i}\;,$$

and the coefficient of $\frac{x^k}{k!}$ in $e^{(n-\ell)x}(e^x-1)^\ell$ is

$$\ell!\sum_{i=\ell}^k\binom{k}i{i\brace\ell}(n-\ell)^{k-i}\;.$$

Get number of functions f from A to B where F(X) $\supseteq$ Y, Y $\subseteq$ B

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