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Let $A = \{1,2,\ldots,m\}$; $B = \{1,2,\ldots,n\}$.

I have to prove that there are at least $\frac{m!}{(m-n+1)!}$ surjective functions from $A$ to $B$.

I've given it some thought, but I don't know how to work out the proof. I've looked at some similar answers, but I'm not finding them helpful. Could anyone please give me a hint or tell me how I need to attack this?

I understand that there are $m!$ ways to arrange the elements of $A$, and that there are $n!$ ways to arrange the elements of $B$. I'm not sure how to draw the possible correspondences between them. I also know that the collection of all $n$-element subsets of $m$ can be expressed as ${m \choose n} = {m \choose m-n}$.

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Newb
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HINT: Suppose that $f:A\to B$ is a surjection. For $k=1,2,\ldots,n-1$ let $$g_f(k)=\min\{a\in A:f(a)=k\}\;.$$ Show that if $f$ and $h$ are surjections from $A$ to $B$, and $$\langle g_f(1),\ldots,g_f(n-1)\rangle\ne\langle g_h(1),\ldots,g_h(n-1)\rangle\;,$$ then $f\ne h$. How many possibilities are there for the $(n-1)$-tuple $\langle g_f(1),\ldots,g_f(n-1)\rangle$?

Brian M. Scott
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  • I'm still not quite getting it - could you please elaborate a little? – Newb Oct 23 '13 at 17:22
  • @Newb: Can you answer the question at the end? – Brian M. Scott Oct 23 '13 at 17:28
  • No. I followed up to that question. It appears to me that there are $(n-1)!$ ways to arrange the tuple itself, but I'm not sure how to relate that back to $A$. – Newb Oct 23 '13 at 17:40
  • @Newb: It’s a sequence of $n-1$ distinct objects chosen from the set $A$, which has altogether $m$ members. How many ways are there to choose such a sequence? This is one of the basic counting tasks. – Brian M. Scott Oct 23 '13 at 17:44
  • Sure, that's ${m \choose n-1} = {m \choose m-n+1} = \frac{m!}{(m-n+1)!(n-1)!}$ But how do I show that $\frac{m!}{(m-n+1)!(n-1)!} \geq \frac{m!}{(m-n+1)!}$? – Newb Oct 23 '13 at 17:51
  • @Newb: No, it isn’t $\binom{m}{n-1}$: that’s the number of $(m-1)$-element subsets of $A$, and we want the number of $(m-1)$-term sequences of distinct elements of $A$ (which is a quantity that actually appears somewhere in your last comment). – Brian M. Scott Oct 23 '13 at 17:53
  • I'm actually not familiar with choosing $x$-term sequences from a set. How do I go about doing that?

    edit: Hold on - there are $m$ choices for the first element, $m-1$ for the second, etc. So it's like a factorial, but I have to stop it at some $m-x$?

     – Newb Oct 23 '13 at 18:06
  • @Newb: No, not $(m-1)!$: there are only $n-1$ of them, so the numbers of choices range from $m-0=m$ down to $m-(n-2)=m-n+2$. Alternatively, you can think of it as choosing a set of $n-1$ things, which can be done in $\binom{m}{n-1}$ ways, and then arranging them in a particular order, which can be done in $(n-1)!$ ways. – Brian M. Scott Oct 23 '13 at 18:12
  • Got it. Finally. Sorry for needing the careful walk-through. Anyway, thank you so much for your help. – Newb Oct 23 '13 at 18:24
  • @Newb: You’re very welcome. Even if it takes a while, I like it better when you make the final step yourself. – Brian M. Scott Oct 23 '13 at 18:25