Compute the following

Question

n! / (n-k-1)!

Answer 1

(n-k-1)! = 1*2*3*...*(n-k-1)
so when you divide n! by that you get
n(n-1)...(n-k)

read up on falling factorials

Answer 2

n! = n(n-1)(n-2)...3*2*1
(n-k-1)! = (n-k-1)(n-k-2)(n-k-3)...*3*2*1

Thus, n!/(n-k-1)! = n(n-1)...(n-k)