If nPr =nCr..... x. ,then what is the value of x?

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Answer 1 · 2018-04-19T02:52:41

When #r=1, x=n#; When #r=0, x=1#

I think #x# is the point where we have #nPr=nCr# so I'll assume that to be the case.

Where we have this relation, we have:

#(n!)/((n-r)!)=(n!)/((k!)(n-r)!)#

We can divide through by #(n!)/((n-k)!)# to get to:

#1=1/(r!)#

#r! =1=>r=1, 0#

Does this work?

Let's have a test case of #n=5, r=1#:

#(5!)/(4!)=(5!)/((1!)(4!))=5#

And now a test case of #n=5, r=0#:

#(5!)/(5!)=(5!)/((5!)(0!))=1#

And so we end up with 2 cases: