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

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Solution

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

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