```Question 117274

<pre><font size = 4 face = "book antiqua"><b>

{{{7P(n,3)}}} = {{{6P(n+1,3)}}}

Not sure why the answer is n=20.

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Use the identity {{{P(k,r)}}} = {{{k!/((k-r)!)}}}
to rewrite the permutations on both sides:

{{{7(n!/(n-3)!)  }}} = {{{ 6((n+1)!/(n+1-3)!)}}}

{{{(7n!)/(n-3)!  }}} = {{{ (6(n+1)!)/(n-2)!)}}}

Multiply both sides by {{{(n-2)!/1}}}

{{{(n-2)!/1}}}{{{(7n!)/(n-3)!  }}} = {{{(n-2)!/1}}}{{{ (6(n+1)!)/(n-2)!)}}}

On the left, write {{{(n-2)!}}} as {{{(n-2)(n-3)!}}}

{{{((n-2)(n-3)!)/1}}}{{{(7n!)/(n-3)!  }}} = {{{(n-2)!/1}}}{{{ (6(n+1)!)/(n-2)!)}}}

Cancel the {{{(n-3)!}}}'s on the left and the
{{{(n-2)!}}}'s on the right:

{{{(n-2)   cross((n-3)!)    )/1}}}{{{(7n!)/(cross(  (n-3)!))  }}} = {{{cross((n-2)!))/1}}}{{{ (6(n+1)!)/cross((n-2)!)))}}}

{{{7(n-2)n!}}} = {{{6(n+1)!}}}

On the right, write {{{(n+1)!}}} as {{{(n+1)n!}}}

{{{7(n-2)n!}}} = {{{6(n+1)n!}}}

Divide both sides by {{{n!}}}

{{{7(n-2)}}} = {{{6(n+1)}}}

{{{7n-14)}}} = {{{6n+6)}}}

{{{n}}} = {{{20}}}

Edwin</pre>

```