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I'm in stuck with this dimostration. I've got $$\frac{n!}{(n+1)!}$$ and it's must be $$\frac{1}{n+1}$$ If I put n=3, I've got $$\frac{3!}{(3+1)!}=\frac{1}{4}=\frac{1}{3+1}$$ and it's correct. But I can't prove the resolution without replacing, like in this case that I put n=3. Can someone explain to me how to do it?

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closed as unclear what you're asking by José Carlos Santos, Cesareo, Hayk, Xander Henderson, metamorphy May 22 at 16:07

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    $\begingroup$ Please fix the typesetting. $\endgroup$ – Yves Daoust May 21 at 10:06
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    $\begingroup$ How do you define the factorial ? $\endgroup$ – Yves Daoust May 21 at 10:07
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    $\begingroup$ What Yves means, is that you can use the definition of factorial (plug it in the expression) and you should be able to proceed. $\endgroup$ – Matti P. May 21 at 10:09
  • $\begingroup$ The definition is n! =n(n-1)! And n! /(n-k)! =n(n-1)(n-k+1) $\endgroup$ – Ciao May 21 at 10:15
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    $\begingroup$ Literally the first thing you wrote, yields the result. $\endgroup$ – maxmilgram May 21 at 10:18
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Recall the definition of factorial, $n! = 1 \times 2 \times \ldots \times n$. So we have $(n+1)! = 1 \times 2 \times \ldots \times n \times (n+1) = n! \times (n+1)$. But from $n! \times (n+1) = (n+1)!$ we simply rearrange to get the desired result.

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  • $\begingroup$ Ok maybe I solved it. n! =n!(n-1) and (n+1)!=n!(n-1)(n+1). I simplify the in fraction n!(n-1). And the rest is 1/(n+1). Is it correct? $\endgroup$ – Ciao May 21 at 11:12
  • $\begingroup$ Almost, you should get n! = n! and (n+1)! = n!(n+1), and cancel out n! inside the fraction $\endgroup$ – auscrypt May 21 at 11:13

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