Question

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

Answer 1

We have,
$(n+1)!=90\left[\right(n-1)!]$
$\Rightarrow (n+1)\times n\times (n-1)!=90\left[\right(n-1\left)\right]!$
$\Rightarrow n(n+1=90)$
$\Rightarrow n^2+n=90$
$\Rightarrow n^2+n-90=0$
$\Rightarrow n^2+10n-9n-90=0$
$\Rightarrow n(n+10)-9(n+10)=0$
$\Rightarrow (n-9)(n+10)=0$
$\Rightarrow n-9=0$ $[\because n+10\ne 0]$
$\Rightarrow n=9$
Hence, n = 9