Obviously if two sets $X$ (with cardinality $m$) and $Y$ (with cardinality $n$) are finite then the number of injective functions from $Y$ to $X$ is $n!/(n-m)!$. However, if $X$ is infinite, what is the correct answer then? I'm inclined to say that it is $n^m$, but I'm not sure how to go about proving this. Any suggestions?
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1Your count is not correct in the finite case. – Tobias Kildetoft Oct 07 '13 at 19:16
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Neither is your iclination for the infinite case. – AlexR Oct 07 '13 at 19:18
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sorry, fixed it – mk_ev112 Oct 07 '13 at 19:18
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@mk_ev112 it's still wrong. – AlexR Oct 07 '13 at 19:20
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it was a rollback error, i think it is good – mk_ev112 Oct 07 '13 at 19:23
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It's $$\binom{m}{n} n!$$ (why?), you had $n^m$... – AlexR Oct 07 '13 at 19:24
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Your $\frac{n!}{(m-n)!}$ would be $\binom{n}mm!$ if you had $n-m$ instead of $m-n$; that’s the number of injective functions from $X$ to $Y$, not $Y$ to $X$. As it stands, it’s a hybrid that doesn’t count much of anything. – Brian M. Scott Oct 07 '13 at 19:25
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@BrianM.Scott except he also has $n$ and $m$ switched in the denominator. – Tobias Kildetoft Oct 07 '13 at 19:26
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This might help http://math.stackexchange.com/questions/492433/if-a-30-and-b-20-find-the-number-of-surjective-functions-fa-to-b/492441#492441 – Rustyn Oct 07 '13 at 19:26
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@Tobias: sigh A case of seeing what one expects to see. – Brian M. Scott Oct 07 '13 at 19:28