Question 155008
{{{P(n,5)=20P(n,3)}}}
<pre><font size = 4 color = "indigo"><b>
{{{P(n,r)=n(n-1)(n-2)}}}···{{{(until_there_are_r_factors)}}}

So, {{{P(n,5)=n(n-1)(n-2)(n-3)(n-4)}}}

and {{{P(n,3)=n(n-1)(n-2)}}}

Substituting,

{{{n(n-1)(n-2)(n-3)(n-4)=20*n(n-1)(n-2)}}}

Get a 0 on the right:

{{{n(n-1)(n-2)(n-3)(n-4)-20*n(n-1)(n-2)=0}}}

factor out {{{n(n-1)(n-2)}}}

{{{n(n-1)(n-2)*((n-3)(n-4)-20)=0}}}

Simplify the last parentheses:

{{{n(n-1)(n-2)((n^2-7n+12)-20)=0}}}

{{{n(n-1)(n-2)(n^2-7n+12-20)=0}}}

{{{n(n-1)(n-2)(n^2-7n-8)=0}}}

Factor the last parentheses:

{{{n(n-1)(n-2)(n-8)(n+1)=0}}}

Setting each factor = 0, we get the
solutions:

{{{n=0}}}, {{{n=1}}}, {{{n=2}}}, {{{n=8}}}, {{{n=-1}}}.

The first three are considered correct,
because when {{{r > n}}}, {{{P(n,r) = 0}}}

And {{{n=8}}} is a solution.  But the last one
{{{n=-1}}} is not a solution because permutations
are not defined for negative numbers.

So the four solutions are

{{{n=0}}}, {{{n=1}}}, {{{n=2}}}, {{{n=8}}}

Possibly your teacher may want only
the solution {{{n=8}}}; however, the first 
three are solutions since they cause 
both sides of the original equation
to equal 0.

Edwin</pre>