Question 934927
<pre>
The permutation formula is 
<font size = 4><b>
xPy = {{{x!/(x-y)!}}}</font></b>

So we are to prove:
<font size = 4><b>
nPr &#8799; (n-1)P(r) + r*((n-1)P(r-1))</font></b>, which becomes:

{{{n!/(n-r)!}}} &#8799; {{{((n-1)!/(n-1-r)!)+r*((n-1)!/((n-1)-(r-1))! )}}}

{{{n!/(n-r)!}}} &#8799; {{{((n-1)!/(n-1-r)!)+r*((n-1)!/(n-1-r+1)! )}}}

{{{n!/(n-r)!}}} &#8799; {{{((n-1)!/(n-1-r)!)+r*((n-1)!/(n-r)!)}}}

--------------------------------------------------------------------
Proof:

nPr = {{{n!/(n-r)!}}}{{{""=""}}}{{{n*(n-1)!/((n-r)(n-r-1)!)}}}{{{""=""}}}{{{(n/(n-r))*((n-1)!/(n-1-r)!)}}}{{{""=""}}}

Now divide {{{n/(n-r)}}} by long division:

              <u>   1</u>                  
           n-r)n+0            
               <u>n-r</u>
                 r

           The quotient is {{{1+r/(n-r)}}}

{{{(1+r/(n-r))*((n-1)!/(n-1-r)!)}}}{{{""=""}}}{{{1*((n-1)!/(n-1-r)!)+(r/(n-r))( (n-1)!/(n-1-r)! )}}}{{{""=""}}}
{{{((n-1)!/(n-1-r)!)+( r(n-1)!/(n-r)(n-1-r)! )}}}{{{""=""}}}{{{((n-1)!/(n-1-r)!)+r*((n-1)!/(n-r)(n-r-1)! )}}}{{{""=""}}}
{{{((n-1)!/(n-1-r)!)+r*((n-1)!/(n-r)! )}}}{{{""=""}}}

<font size = 4><b>
nPr = (n-1)P(r) + r*[(n-1)P(r-1)]</font></b>
 
Edwin</pre>