Question

A7,2+ A4,3? Por favor

Answer 1

$\implies\boxed{A_{n,p}=\dfrac{n!}{(n-p)!}}$

$\iff A_{7,2}+A_{4,3}=\dfrac{7!}{(7-2)!}+\dfrac{4!}{(4-3)!}$

${\iff A_{7,2}+A_{4,3}=\dfrac{7\cdot6\cdot\diagup\!\!\!\!5!}{\diagup\!\!\!\!5!}+\dfrac{4\cdot3\cdot2\cdot\diagup\!\!\!\!1!}{\diagup\!\!\!\!1!} }$

$\iff A_{7,2}+A_{4,3}= 42+ 24$

$\iff\red{\boldsymbol{A_{7,2}+A_{4,3}=66 }}$