How do you simplify $\frac{(2 k - 2)!}{2 k!}$ $((2k-2)!) /( 2k!)$ ?

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How do you simplify $\frac{(2 k - 2)!}{2 k!}$ $((2k-2)!) /( 2k!)$ ?

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Answer 1 · 2015-11-10T18:52:40

I found: $\frac{1}{2 k (2 k - 1)}$ $1/(2k(2k-1))$

Explanation:

I tried considering that:
$k! = k \cdot (k - 1)!$ $k! =k*(k-1)!$
Rearranging your expression I got:
$((2k-2)!)/(2k!)=(cancel((2k-2))cancel((2k-3)!))/(2k(2k-1)cancel((2k-2))cancel((2k-3)!))=1/(2k(2k-1))$
I tried with $k=5$
$((2k-2)!)/(2k!)=(8!)/(10!)=0.0bar1$
and:
$1/(2k(2k-1))=1/(10*9)=0.0bar1$

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How do you simplify $\frac{(2 k - 2)!}{2 k!}$ $((2k-2)!) /( 2k!)$ ?

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How do you simplify $\frac{(2 k - 2)!}{2 k!}$ $((2k-2)!) /( 2k!)$ ?