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

Question:

If (n + 3)! = 56 [(n + 1)!], find n.

Solution:

(n + 3)! = 56 [(n + 1)!]

$\Rightarrow(n+3) \times(n+2) \times(n+1) !=56[(n+1) !]$

 

$\Rightarrow(n+3) \times(n+2)=56$

$\Rightarrow(n+3) \times(n+2)=8 \times 7$

$\Rightarrow n+3=8$

 

$\therefore n=5$

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