Da fórmula de Combinação [tex]\boxed{C_{n,p}=\frac{n!}{(n-p)!p!}}[/tex], temos que:
[tex]C_{n,p}=\frac{n!}{(n-p)!p!}\Leftrightarrow C_{n,2}=\frac{n!}{(n-2)!2!}\\\\\\n+2=\frac{n(n-1)(n-2)!}{(n-2)!2\cdot1}\\\\\\\frac{n(n-1)}{2}=n+2\\\\n(n-1)=2(n+2)\\\\n^2-n=2n+4\\\\n^2-3n-4=0\\\\n^2-4n+n-4=0\\\\n(n-4)+1(n-4)=0\\\\(n-4)[n+1]=0[/tex]
Uma vez que [tex]n\geq0[/tex], isto é, [tex]n\in\mathbb{N}[/tex] temos que [tex]\boxed{\boxed{n=4}}[/tex]
