Question

CALCULE POR FAVOR AJUDEM!

$a) A^{2} _{7}        b)A^{3} _{9}$

Answer 1

$A_7^2 = \dfrac{7!}{(7-2)!} = \dfrac{7!}{5!} = \dfrac{7.6.\not5!}{\not5!} = 7.6 = \boxed{42} \\ \\ \\ A_9^3=\dfrac{9!}{(9-3)!} = \dfrac{9!}{6!} = \dfrac{9.8.7.\not6!}{\not6!}= 9.8.7 = \boxed{504}$