Evaluate #((n-6)!)/((n-5)!)#?

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Answer 1 · 2017-11-24T00:36:42

#((n-6)!)/((n-5)!)=((n-6)!)/((n-5)(n-6)!)=1/(n-5)#

Remember that #n! = n xx (n-1) xx (n-2) xx ... xx 1#

And so we can look at #(n-5)! = (n-5) xx (n-6) xx ... xx 1#

and #(n-6)! = (n-6) xx (n-7) xx ... xx 1#

Which means that:

#(n-5)! = (n-5) xx (n-6)!#

And now rewrite:

#((n-6)!)/((n-5)!)=((n-6)!)/((n-5)(n-6)!)=1/(n-5)#