Question

calcule o valor de n na igualdade Cn,2= n +2

Answer 1

Veja que: Cn,2 = n!/(n-2)!2!
Cn,2 = n(n-1)/2

Logo: n(n-1)/2 = n+2
n² - n = 2n + 4

n² - 3n - 4 = 0

n = 4 ou n = -1

Como n deve ser natural, teremos n = 4.

Answer 2

Da fórmula de Combinação $\boxed{C_{n,p}=\frac{n!}{(n-p)!p!}}$, temos que:

$C_{n,p}=\frac{n!}{(n-p)!p!}\Leftrightarrow C_{n,2}=\frac{n!}{(n-2)!2!}\\\\\\n+2=\frac{n(n-1)(n-2)!}{(n-2)!2\cdot1}\\\\\\\frac{n(n-1)}{2}=n+2\\\\n(n-1)=2(n+2)\\\\n^2-n=2n+4\\\\n^2-3n-4=0\\\\n^2-4n+n-4=0\\\\n(n-4)+1(n-4)=0\\\\(n-4)[n+1]=0$

 Uma vez que $n\geq0$, isto é, $n\in\mathbb{N}$ temos que $\boxed{\boxed{n=4}}$