n permutation 2,divided by n combination 3

Answer 1

P(n,2) / C(n,3)
= [ n!/(n-2)!] / [ n!/(3!(n-3)!]
= n!/(n-2)! * 3!(n-3)!/n!
= 3!/(n-2)
= 6/(n-2)

Answer 2

or, if you're not comfortable manipulating factorials,
nP2 = n(n-1)
nC3 = n(n-1)(n-2) / (1*2*3)
now divide to get

n(n-1) * (1*2*3) / n(n-1)(n-1) = 6/(n-2)