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 + 10n - 9n - 90 = 0 $
$\Rightarrow n^2 (n + 10) - 9 (n + 10) = 0$
$ \Rightarrow (n^2- 9) (n + 10) = 0$
$\big[\therefore \text{n} +10 \neq 0\big]$
$\Rightarrow n - 9 = 0 $
$\Rightarrow n = 9$
Hence, $n = 9$

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