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: