What is ((n-2)!)/((n+1)!) ?

The solution to ((n-2)!)/((n+1)!) is 1/((n-1)n(n+1))

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simplify ((n-2)!)/((n+1)!)

simplify

\frac{( n - 2 ) !}{( n + 1 ) !}

\frac{1}{( n - 1 ) n ( n + 1 )}

Solution steps

\frac{( n - 2 ) !}{( n + 1 ) !}

Cancel the factorials:

\frac{n !}{( n + m ) !} = \frac{1}{( n + 1 ) \cdot ( n + 2 ) \dots ( n + m )}

\frac{( n - 2 ) !}{( n + 1 ) !} = \frac{1}{( n - 1 ) n ( n + 1 )}

= \frac{1}{( n - 1 ) n ( n + 1 )}

8 x^{3} + y^{6}

(9 c^{- 9})^{- 3}

a^{- 1} + 1

\frac{9 x ^{2} - 1}{6 x ^{2} + 5 x + 1}

(\frac{1 - 1}{x ^{2}})^{- 1}

What is ((n-2)!)/((n+1)!) ?
The solution to ((n-2)!)/((n+1)!) is 1/((n-1)n(n+1))