Question 144803
25. solve for n if {{{P(n,3)=2C(n,2)}}}, n € N
<pre><font size = 4 color="indigo"><b>
{{{P(N,R) = N*(N-1)*(n-2)}}}··· (until there are R factors)


{{{C(N,R) = (N*(N-1)*(N-2)_until_there_are_R_factors)/(R*(R-1)*(R-2)_until_you_get_to_1)}}}


So

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

and

{{{C(n,2)=n(n-1)/(2*1)}}}

So

{{{P(n,3)=2C(n,2)}}} 

becomes:

{{{n(n-1)(n-2)=2(n(n-1)/2)}}}

{{{n(n-1)(n-2)=cross(2)(n(n-1)/cross(2))}}}

{{{n(n-1)(n-2)=n(n-1)}}}

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

Factor out {{{n(n-1)}}} on left side:

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

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

{{{n(n-1)(n-3)=0}}}

Set each factor = 0

{{{n=0}}}

{{{n-1=0}}} or {{{n=1}}}

{{{n-3=0}}} or {{{n=3}}}

So there are three solutions, {{{n=0}}}, {{{n=1}}}, {{{n=3}}}

Edwin</pre>