Question

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

Answer 1

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

Explanation:

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:

• When `r=1, x=n`
• When `r=0, x=1`