If (n + 1)! = 90 [(n - 1)!], find n. - Vidyadip

Question

If $(n + 1)! = 90 [(n - 1)!]$, find $n$.

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Answer

We have, $(n+1)!=90[(n-1)!]$
$\Rightarrow(n+1)!\times n \times(n-1)!=90[(n-1)!]$
$\Rightarrow n(n+1)=90$
$\Rightarrow n^2+10 n-9 n-90=0$
$\Rightarrow n^2(n+10)-9(n+10)=0$
$\Rightarrow\left(n^2\right.$
$-9)(n+10)=0[\therefore n+10 \neq 0]$
$\Rightarrow n-9=0$
$\Rightarrow n=9 \text { Hence, } n=9$

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