If (n+3)! = 56 [(n-1)!] find n.

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Solution

We have,
$(n + 3)! = 56 [(n - 1)!]$
$\Rightarrow (n + 3) \times (n + 2) \times (n + 1)!$
= $56 [(n + 1)!]$
$\Rightarrow (n + 2) (n + 3) = 56$
$\Rightarrow n^{2} + 3 n + 2 n + 6 = 56$
$\Rightarrow n^{2} + 5 n + 6 - 56 = 0$
$\Rightarrow n^{2} + 5 n - 50 = 0$
$\Rightarrow n^{2} + 10 n - 5 n - 50 = 0$
$\Rightarrow n (n + 10) - 5 (n + 10) = 0$
$\Rightarrow (n + 10) (n - 5) = 0$
$\Rightarrow n - 5 = 0$ $[∵ n + 10 \neq 0]$
$\Rightarrow n - 5 = 0$
$\Rightarrow n = 5$

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