How do you simplify #((n-k)!) /( n!)#?

1 Answer
Topscooter
Dec 18, 2015

#((n-k)!)/(n!) = 1/((n-k+1)!)#

Explanation:

You simply develop #n!# and #(n-k)!#. #n-k < n# so #(n-k)! < n!# and #(n-k)!# divides #n!#.

All the terms of #(n-k)!# are included in #n!#, hence the answer.

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