Question 847247
nC(n-2)  =   {{{n!/(n-2)!(2!) = 10}}}

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

So we have that  {{{(n*(n-1))/2 = 10}}}

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

{{{n^2 -n = 20}}}

{{{n^2 -n -20 = 0}}}

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

n = 5.