If (n + 1)! = 90 [(n − 1)!], find n.
Question:

If (n + 1)! = 90 [(n − 1)!], find n.

Solution:

(n + 1)! = 90 [(n − 1)!]

$\Rightarrow(n+1) \times(n) \times(n-1) !=90[(n-1) !]$

$\Rightarrow(n+1) \times(n)=90$

 

$\Rightarrow(n+1) \times(n)=10 \times 9$

On comparing, we get:

n = 9

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