Question

Resolva a equação binomial:

C ( n+2, 4 ) = 11. C ( n, 2 )

Answer 1

$\boxed{C_{n,k}=\dfrac{n!}{k!\cdot \left(n-k \right )!}}$


$C_{n+2,\;\;4}=11 \cdot C_{n,\;\;2}\\ \\ \dfrac{\left(n+2 \right )!}{4!\cdot \left(n+2-4 \right )!}=11\cdot\dfrac{n!}{2!\cdot \left(n-2 \right )!}\\ \\ \dfrac{\left(n+2 \right )!}{4!\cdot \left(n-2 \right )!}=11\cdot\dfrac{n!}{2!\cdot \left(n-2 \right )!}$


Multiplicando os dois lados por $\left(n-2\right)!$, para cancelar este fator nos denominadores, temos

$\dfrac{\left(n+2 \right )!}{4!}=11\cdot\dfrac{n!}{2!}\\ \\ \dfrac{\left(n+2 \right )\cdot \left(n+1 \right )\cdot n!}{4 \cdot 3 \cdot 2!}=11\cdot\dfrac{n!}{2!}\\ \\ \dfrac{\left(n+2 \right )\cdot \left(n+1 \right )}{4 \cdot 3}\cdot \dfrac{n!}{2!}=11\cdot\dfrac{n!}{2!}\\ \\ \dfrac{\left(n+2 \right )\cdot \left(n+1 \right )}{12}=11\\ \\ \left(n+2 \right )\cdot \left(n+1\right)=11\cdot 12$

Inspecionando a equação, podemos verificar que

$n+1=11\\\ n+2=12$


Logo, 

$\boxed{n=10}$