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

Question

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

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Answer

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

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