Question #24247

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Answer 1 · 2015-03-30T00:19:04

$P(a,b) = (a!)/((a-b)!)$
where $a! = a xx (a-1) xx(a-2)xx(a-3)xx ....xx3 xx 2 1$
and $(a-b)! = (a-b) xx (a-b-1) xx (a-b-2)xx .... xx 2 xx1$

So
$P(n,r) =(n!)/((n-r)!)$
and
$P(n-1,r-1) = ((n-1)!)/(((n-1)-(r-1))!) = ((n-1)!)/((n-r)!)$
so

$n*P(n-1,r-1)$

$= ((n)(n-1)!)/((n-r)!)$

$= (n xx (n-1) xx (n-2) xx ....xx2 xx1)/((n-r)!)$

$= (n!)/((n-r)!)$

$=P(n,r)$