Find $n$ , if :
(i) $(n + 1)! = 42 (n - 1)!$
(ii) $(n + 3)! = 110 (n + 1)!$

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Solution

(i) $(n + 1)! = 42 (n - 1)!$
$(n + 1) (n) (n - 1)! = 42 (n - 1)!$
$n^{2} + n = 42$
$n^{2} + n - 42 = 0$
$n^{2} + 7 n - 6 n - 42 = 0$
$n (n + 7) - 6 (n + 7) = 0$
$(n - 6) (n + 7) = 0$
[n=6] $[n \neq - 7]$
Hence [n=6]

(ii) $(n + 3)! = 110 (n + 1)!$
$\Rightarrow (n + 3) (n + 2) (n + 1)! = 110 (n + 1)!$
$\Rightarrow n^{2} + 5 n + 6 - 110 = 0$
$\Rightarrow n^{2} + 5 n - 104 = 0$
$\Rightarrow n^{2} + 13 n - 8 n - 104 = 0$
$\Rightarrow n (n + 3) - 8 (n + 13) = 0$
$\Rightarrow (n - 8) (n + 13) = 0$
$\Rightarrow [n = 8] [n \neq - 13]$

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