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I'm having trouble on this question:

Let $f(n,r)$ be the number of surjections from a set $A$ having $n$ elements to a set $B$ having $r$ elements. Show that $$f(n,r)=r\Big(f(n-1,r-1)+f(n-1,r)\Big)\;.$$

Here is my idea about how to start:

Partition each set, $A$ and $B$, such that the top partition consists of $n-1$ or $r-1$ elements (for $A$ and $B$ respectively) and the bottom partitions consists of one element each.

Then there are $f(n-1,r-1)$ surjections from the top partition of $A$ onto the top partition of $B$.

There are $f(n-1,r)$ surjections from the top partition of $A$ to all of $B$.

Now consider the whole of $A$ (i.e. $(n-1)+1$ elements).

The total number of surjections is:

((total number of surjections from top partition of $A$ onto all of $B$) + (extra surjections due to extra element of $A$)) permuted to account for all combinations

But how do you calculate the extra surjections due to the extra element of $A$ and the correct number of permutations?

Thank you.

Brian M. Scott
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Froskoy
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    The formula for $f(n,r)$ can't be right: $f$ is a function of two arguments, but the outer $f$ on the righthand side has only one. – Brian M. Scott Apr 30 '12 at 14:30
  • Thanks - it was a typo - I've corrected it. – Froskoy Apr 30 '12 at 15:06
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    You accidentally changed the wrong $f$, but I’ve fixed it now. – Brian M. Scott Apr 30 '12 at 15:14
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    There is a natural correspondence between surjective functions and partitions, so this is the same thing as counting ordered partitions of a given set $A$ into $r$ non-empty subsets. The number of such partitions was used in this paper: Gupta, H.N. (1985): A theorem in combinatorics and Wilson's theorem. Amer. Math. Monthly, 92, 575–576. jstor link. If we omit the word ordered, we get Stirling numbers of the second kind and, indeed, for these numbers we can get very similar formula. – Martin Sleziak Apr 30 '12 at 15:22

2 Answers2

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I'm going to go out on a limb and suppose the intended identity is $$ f(n,r) = r(f(n-1,r-1) + f(n-1,r)). $$

For convenience, let $A = \{1, \dots, n\}$ and $B = \{1, \dots, r\}$.

Let's call our sujection $g$. First, let's decide where to map the element 1 under $g$. There are clearly $r$ places it could go.

Now, one of two things could happen. Either $g(1) \neq g(j)$ for all $j \neq 1$ or not.

In the former case, there are $f(n-1,r-1)$ possible mappings, since we are required to find a surjection from $A - \{1\}$ onto $B - \{g(1)\}$.

In the latter case, there are $f(n-1,r)$ possible mappings, since we are required to find a surjection from $A - \{1\}$ onto $B$ (remember, in this case we are specifically requiring that something else gets mapped to $g(1)$).

Putting all this together gives the desired identity.

Austin Mohr
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Assuming that the formula is supposed to be $$f(n,r)=r\Big(f(n-1,r-1)+f(n-1,r)\Big)\;,$$ something similar to your basic idea can be made to work.

Fix $a\in A$, let $A'=A\setminus\{a\}$, and suppose that $\varphi:A\to B$ is a surjection. There are two possibilities: either $\varphi[A']=B$, so that $\varphi\upharpoonright A'$ is a surjection of $A'$ onto $B$, or there is exactly one $b\in B\setminus\varphi[B]$, so that $\varphi\upharpoonright A'$ is a surjection of $A'$ onto $B\setminus\{b\}$. In the first case there is no restriction on $\varphi(a)$: it can be any member of $B$. In the second case, however, we must have $\varphi(a)=b$, since $\varphi$ is a surjection. Let $\Phi_1$ be the set of surjections of $A$ onto $B$ covered by the first case and $\Phi_2$ the set covered by the second case.

There are $f(n-1,r)$ surjections of $A'$ onto $B$. If $\psi$ is one of these surjections, there are $r$ surjections $\varphi\in\Phi_1$ such that $\varphi\upharpoonright A'=\psi$, one for each of the $r$ possible values of $\varphi(a)$. Thus, $|\Phi_1|=rf(n-1,r)$.

For each $b\in B$ there are $f(n-1,r-1)$ surjections of $A'$ onto $B\setminus\{b\}$, and there are $r$ possible choices of $b$, so $|\Phi_2|=rf(n-1,r-1)$.

The sets $\Phi_1$ and $\Phi_2$ are disjoint and exhaust the set of surjections of $A$ onto $B$, so $$f(n,r)=r\Big(f(n-1,r-1)+f(n-1,r)\Big)\;.$$

Brian M. Scott
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