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How to prove that $n! = n^n - C_{n,1} (n-1)^n +C_{n,2} (n-2)^n - \cdots\,{} $?

I faced this problem when trying to find the number of onto functions possible from one set having n elements to another set having $n$ elements.

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  • What are the $C_i$? – Akiva Weinberger Apr 14 '16 at 15:18
  • It's the combinations symbol https://en.m.wikipedia.org/wiki/Combination. It's an alternative way to write n choose r. @AkivaWeinberger –  Apr 14 '16 at 15:20
  • It is true for $n=1$. If you assume it for $n$, can you prove it for $n+1$? – Colin McLarty Apr 14 '16 at 15:21
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    That would make sense if there were two numbers, such as $C_r^n$ or $C_{n,r}$. – Akiva Weinberger Apr 14 '16 at 15:22
  • @ColinMcLarty I was looking for a derivation rather than a proof by induction –  Apr 14 '16 at 15:23
  • @SanchayanDutta: would you accept a derivation by induction ? –  Apr 14 '16 at 15:24
  • Well I didn't know that's possible.How ? @YvesDaoust –  Apr 14 '16 at 15:26
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    You may recognize that the expression on the right has an Inclusion/Exclusion shape. The $n^n$ counts all functions. If $a$ is any of $1,2,\dots,n$, then $(n-1)^n$ counts the funtions that never take on the value $a$. This $a$ can be chosen in $\binom{n}{1}$ ways, giving, sort of, a total of $\binom{n}{1}(n-1)^n$. – André Nicolas Apr 14 '16 at 15:27
  • I got that @AndréNicolas but how to get the n! on the left ? –  Apr 14 '16 at 15:30
  • Ok I finally understood the inclusion exclusion you meant.Thanks a lot –  Apr 14 '16 at 15:34
  • I don't understand why this question is downvoted: this is a very interesting question. The only advice we can strongly give to the OP is, indeed, to replace his $C_k$ which is completely ambiguous notation, by $C_n^r$ or the now international $\binom{n}{r}$. – Jean Marie Apr 14 '16 at 15:34
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    @SanchayanDutta: I assume you got it, but to make sure, the $n!$ counts the number of bijective functions from an $n$-element set to (say) itself. These are just the onto functions. Then the right-hand side counts these in a different way, by looking at all $n^n$ functions, and adjusting by removing the "bad" ones. – André Nicolas Apr 14 '16 at 15:46
  • @JeanMarie I did not downvote, but another piece of advice would be: do not ask for a proof if you will not accept a proof by induction. If you only want a bijective proof, then ask for a bijective proof. – Colin McLarty Apr 14 '16 at 18:56

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Let $(e^x-1)^n$ expanded by the binomial theorem.After expanding it we will get

$(e^x-1)^n=e^{nx}-{{n}\choose{1}}e^{(n-1)x}+...+(-1)^n$. We know that

$e^x=1+x+x^2/2!+x^3/3!+...+x^n/n!+...$

Now we have $(1+x+x^2/2!+...+x^n/n!+...-1)^n=[1+nx+(nx)^2/2!+...+(nx)^n/n!+...]-{{n}\choose{1}}[1+(n-1)x+{(n-1)^2x}^2/2!+...+{(n-1)^nx}^n/n!+...]+...+(-1)^{(n-1)}{{n}\choose{n-1}}[1+x+x^2/2!+...+x^n/n!+...]+(-1)^n$

Now comparing coefficient of x^n on both sides we get

$$1=n^n/n!-{{n}\choose{1}}(n-1)^n/n!+{{n}\choose{2}}(n-2)^n/n!-...+(-1)^n{{n}\choose{n-1}}/n!$$

Now multiply both sides by $n!$, then you will get answer for your question.

sayan
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