P(n,k) = n!/(n-k)!
then ... P(n, 2) = n!/(n-2)! and P(n+1, 2) = (n+1)!/(n+1 - 2)! = (n+1)!/(n-1)!
so ...
P(n, 2)/P(n+1, 2) = [n!/(n-2)!]/[(n+1)!/(n-1)!] = [n!/(n-2)!]*[(n-1)!/(n+1)!]
now, note that (n-1)! = (n-1)(n-2)! and (n+1)! = (n+1)n!
using the two equations given above, you should be able to simplify the expression [n!/(n-2)!]*[(n-1)!/(n+1)!]