Question

1. Вычислите: a) A b) C3 c) Pa Решите​

Answer 1

Объяснение:

$\displaystyle\\a)\\\\A_7^4=\frac{7!}{(7-4)!}=\frac{7!}{3!} =\frac{3!*4*5*6*7}{3!}=4*5*6*7=840.\\\\ b)\\\\C_7^3=\frac{7!}{(7-3)!*3!}=\frac{4!*5*6*7}{4!*1*2*3}=\frac{5*6*7}{6}=5*7=35.\\\\ c)\\\\\frac{P_9}{P_{11}} =\frac{9!}{11!} =\frac{9!}{9!*10*11} =\frac{1}{110}.$