Question

Simplifique as expressões de Análise Combinatória.

Answer 1

$\large\boxed{\begin{array}{l}\sf a )\\\rm C_{6,4}=\dfrac{6\cdot5\cdot4\cdot3}{4\cdot3\cdot2\cdot1}=\dfrac{360}{24}=15\\\\\rm A_{6,4}=6\cdot5\cdot4\cdot3=360\\\\\rm \dfrac{C_{6,4}}{A_{6,4}}\cdot6!=\dfrac{15}{360}\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\\\\\rm \dfrac{C_{6,4}}{A_{6,4}}\cdot6!=\dfrac{10800}{360}=30\end{array}}$

$\large\boxed{\begin{array}{l}\sf b)\\\rm C_{2,0}=\dfrac{2!}{0!\cdot(2-0)!}=\dfrac{\diagdown\!\!\!\!\!2!}{1\cdot\diagdown\!\!\!\!\!\!2!}=1\\\\\rm A_{5,1}=\dfrac{5!}{(5-1)!}=\dfrac{5!}{4!}=\dfrac{5\cdot4!}{4!}=5\\\\\rm C_{2,0}-\dfrac{1}{A_{5,1}}=1-\dfrac{1}{5}=\dfrac{5-1}{5}=\dfrac{4}{5}\end{array}}$

$\large\boxed{\begin{array}{l}\sf c)\\\rm\dfrac{C_{n,p}}{P_n}=C_{n,p}\div P_n\\\rm C_{n,p}=\dfrac{n!}{p!\cdot(n-p)!}\\\\\rm P_n=n!\\\\\rm C_{n,p}\div P_n=\bigg(\dfrac{n!}{p!\cdot(n-p)!}\bigg)\div n!\\\\\rm C_{n,p}\div P_n=\bigg(\dfrac{\diagdown\!\!\!\!\!\!n!}{p!\cdot(n-p)!}\bigg)\cdot\dfrac{1}{\diagdown\!\!\!\!\!n!}\\\\\rm\dfrac{C_{n,p}}{P_n}=\dfrac{1}{p!\cdot(n-p)!}\end{array}}$

$\large\boxed{\begin{array}{l}\sf d)\\\rm\dfrac{P_{n+1}}{n+1}=\dfrac{(n+1)!}{n+1}=\dfrac{\diagup\!\!\!\!(n+\diagup\!\!\!\!1)\cdot n!}{\diagup\!\!\!\!\!(n+\diagup\!\!\!\!1)}\\\\\rm\dfrac{P_{n+1}}{n+1}=n!\end{array}}$