Question 684256
 	

when you have a question like {{{nC2=10}}}
to solve for {{{n}}}, you can do it this way:


{{{n(n-1)(n-2)/2!(n-2)! = 10}}}

{{{n(n-1)/2 = 10}}}

{{{n(n-1) = 20}}}

{{{n^2-n-20=0}}}....replace {{{-n}}} with {{{-5n+4n}}} and factor

{{{n^2-5n+4n-20=0}}}

{{{(n+4)(n-5)=0}}}

solutions:


if {{{(n+4)=0}}} than {{{n=-4}}} or 

if {{{(n-5)=0}}} than {{{n=5}}} ......since {{{n}}} can’t be negative we use only solution {{{n=5}}}



check:

{{{nCr=n!/((n-2)!*r!)}}}

given {{{r=2}}}

solve the expression  {{{nC2=n!/((n-2)!*2!)}}}


plug in {{{n=5 }}} 


 {{{5C2=5!/((5-2)!*2!)}}}


 {{{5C2=(5*4*3*2*1)/(3!*2!)}}}


 {{{5C2=(5*4*3*2*1)/(3*2*1*2*1)}}}


 {{{5C2=(5*cross(4)2*cross(3*2*1))/(cross(3*2*1)*cross(2)*1)}}}


 {{{5C2=5*2}}}


{{{5C2=10}}}