If (n+2)!= 60 [(n-1)!], find n.

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Solution

(n+2)! = 60 [(n-1)!]
(n+2)(n+1)(n) (n-1)! = 60 [(n-1)!]
$\Rightarrow (n + 2) (n + 1) n = 60$
$\Rightarrow (n + 2) (n + 1) n = 5 \times 4 \times 3$
$∴$ n=3 [By comparing]
Hence, n=3

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