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

Question

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

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Answer

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

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