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We have $r$ objects and $n$ boxes. I have to count all the combinations possible if the objects and boxes could be both different. If that happens I can count the number of variations from $n$ to $r$ (number of functions between objects and boxes). I mean, $n^r$.

But if I have to count all the combinations possible if anyone of the boxes is empty ($r\ge n$). Now I have to count the surjective functions.

For example if I have $r$ objects and 1 box. The number of surjective functions is 1. Then, if I have $r$ objects and 2 boxes, the number of surjective functions is $2^r-2$. And finally if I have $r$ objects and 3 boxes, I will count $3^r-2\cdot[3\cdot2^r-3]$. How do I get the formula for $r$ objects and $n$ boxes?

2 Answers2

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We want the number of surjections from $r$ to $n$.

First, note that the Stirling number of the second kind $$ \begin{Bmatrix} x\\ y \end{Bmatrix} $$ is a quantity that solves most of the puzzle: this number equals the number of partitions of $x$ into $y$-many nonempty subsets.

Clearly, every surjection $f\colon [r]\to [n]$ determines such a partition: $\{f^{-1}(j)\mid 1\le j\le n\}$, and each such partition of $r$ corresponds to $n!$ different surjections (permutations of the range). (Here, $[m]$ denotes $\{1,\dotsc,m\}$.) So the total number of surjections $[r]\to [n]$ is: $$ n! \begin{Bmatrix} r\\ n \end{Bmatrix} $$

The Stirling number of the second kind can be computed from the explicit formula $$ \begin{equation*} \begin{Bmatrix} r\\ n \end{Bmatrix} =\frac{1}{n!}\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}j^r \end{equation*} $$ As you can see, a kind of "inclusion/exclusion" principle is at work. Various answers to the following question (the special case $n=3$ of yours) provide some intuition about why that is so: Counting the number of surjections.

Notice the factor of $\frac 1 {n!}$, which we multiplied the Stirling number of 2nd kind by when counting surjections. So finally can simplify further: the number of surjections $[r]\to [n]$ is $$ \sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}j^r $$

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BrianO
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  • Don't we have to substract for the totsal number of functions the non-surjective ones? –  Nov 12 '15 at 22:58
  • Yes we do, and it does: as $j$ varies from $0$ to $n$, the sign of $(−1)^{n−j}$ keeps alternating. We start with $j=n$ and the summation term $n^r$, which counts all functions from $[r]$ to $[n]$; but that's too big, so have to subtract those which omit at least $1$ point of the range; but then we've gone too far and have to add back those that we just counted multiple times when subtracting, etc. – BrianO Nov 12 '15 at 23:14
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Lets say you have $k$ empty boxes where $1\leq k \leq n-1$. Choose your empty boxes $\binom{n}{k}$ and then fill the other boxes with your objects $(n-k)^r$. Sum this up and you get $\sum_{k=1}^{n-1}\binom{n}{k}(n-k)^r$. The number of your surjective functions will be $n^r-\sum_{k=1}^{n-1}\binom{n}{k}(n-k)^r$.

Alex Fish
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  • And can you do it without $k$? @AlexFish –  Nov 12 '15 at 22:44
  • You can do it for every $r$ but it's not really nice. You can see my sollution here: http://math.stackexchange.com/questions/1526442/prove-binomial-equation/1526467#1526467 for the case $r=2$. You can do the same for any $r$. I can't think of anything better right now. – Alex Fish Nov 12 '15 at 22:48
  • And using $n^r - [n*(n-1)^r - \sum_{k=1}^{ n-1}(n-2)^r]$ ? @AlexFish –  Nov 12 '15 at 22:52