Question

4. Simplifique: (Questão de fatoração)​​

Answer 1

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$\sf\dfrac{C_{n-1,p-1}}{C_{n-2,p-2}}\\\sf C_{n-1,p-1}=\dfrac{(n-1)!}{(p-1)!\cdot(n-1-[p-1])!}\\\sf C_{n-1,p-1}=\dfrac{(n-1)!}{(p-1)!\cdot(n-\diagup\!\!\!1-p+\diagup\!\!\!1)}\\\boxed{\sf C_{n-1,p-1}=\dfrac{(n-1)!}{(p-1)!\cdot(n-p)!}}$

$\sf C_{n-2,p-2}=\dfrac{(n-2)!}{(p-2)!\cdot(n-2-[p-2])!}\\\sf C_{n-2,p-2}=\dfrac{(n-2)!}{(p-2)!\cdot(n-\diagup\!\!\!2-p+\diagup\!\!\!2)!}\\\boxed{\sf C_{n-2,p-2}=\dfrac{(n-2)!}{(p-2)!\cdot(n-p)!}}$

$\sf\dfrac{C_{n-1,p-1}}{C_{n-2,p-2}}=\dfrac{\frac{(n-1)!}{(p-1)!\cdot(n-p)!}}{\frac{(n-2)!}{(p-2)!\cdot(n-p)!}}\\\sf\dfrac{C_{n-1,p-1}}{C_{n-2,p-2}}=\dfrac{(n-1)!}{(p-1)!\cdot\diagup\!\!\!\!(n-\diagup\!\!\!\!p)!}\cdot\dfrac{(p-2)!\cdot\diagup\!\!\!\!(n-\diagup\!\!\!\!p)!}{(n-2)!}$

$\sf \dfrac{C_{n-1,p-1}}{C_{n-2,p-2}}=\dfrac{(n-1)\cdot\diagup\!\!\!\!(n-\diagup\!\!\!\!2)!}{(p-1)\cdot\diagup\!\!\!\!(p-\diagup\!\!\!\!2)!}\cdot\dfrac{\diagup\!\!\!\!(p-\diagup\!\!\!\!2)!}{\diagup\!\!\!\!(n-\diagup\!\!\!\!2)!}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf\dfrac{C_{n-1,p-1}}{C_{n-2,p-2}}=\dfrac{n-1}{p-1}\checkmark}}}}$