Question

How do you simplify `((2k-2)!) /( 2k!)`?

Answer 1

I found: `1/(2k(2k-1))`

Explanation:

I tried considering that:
`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`