Question

Simplifique:

n!/2!(n-2)!
(n+2)!/3!(n-1)!
(2n)!/2!(2n–2)! n.(n-2)!/(n+1)!

Answer 1

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$\boxed{\begin{array}{l}\sf\dfrac{n!}{2!(n-2)!}=\dfrac{n(n-1)\diagdown\!\!\!\!\!(n-\diagdown\!\!\!\!\!\!2)!}{2\diagdown\!\!\!\!\!\!(n-\diagdown\!\!\!\!\!\!2)!}=\dfrac{n(n-1)}{2}\\\sf\dfrac{(n+2)!}{3!(n-1)!}=\dfrac{(n+2)(n+1)n\diagdown\!\!\!\!\!\!(n-\diagdown\!\!\!\!\!\!1)!}{6\diagdown\!\!\!\!\!\!(n-\diagdown\!\!\!\!\!\!1)!}=\dfrac{n(n+2)(n+1)}{6}\end{array}}$

$\boxed{\begin{array}{l}\sf\dfrac{(2n)!}{2!(2n-2)!}=\dfrac{(2n)(2n-1)\diagdown\!\!\!\!\!\!(2n-\diagdown\!\!\!\!\!\!2)!}{2\diagdown\!\!\!\!\!\!(2n-\diagdown\!\!\!\!\!\!2)!}\\\sf=\dfrac{\backslash\!\!\!2n(2n-1)}{\backslash\!\!\!2}=n(2n-1)\\\sf\dfrac{n(n-2)!}{(n+1)!}=\dfrac{n\diagdown\!\!\!\!\!\!(n-\diagdown\!\!\!\!\!\!\!1)!}{(n+1)n\diagdown\!\!\!\!\!\!(n-\diagdown\!\!\!\!\!\!1)!}=\dfrac{\backslash\!\!\!n}{\backslash\!\!\!n(n+1)}\\\sf=\dfrac{1}{n+1}\end{array}}$