Question

помогите решить пожалуйста упражнения 329,330,331,332​

Answer 1

Ответ:

Объяснение:

$329.\\1) A_{3} ^{1} =\frac{3!}{(3-1)!} =\frac{3!}{2!} =\frac{1*2*3}{1*2} =3.\\2)A_{3} ^{2} =\frac{3!}{(3-2)!} =\frac{3!}{1!} =\frac{1*2*3}{1} =6.\\3)A^{2}_{7} =\frac{7!}{(7-2)!} =\frac{7!}{5!} =\frac{5!*6*7}{5!} =6*7=42.\\4)A^{7}_{7} } =\frac{7!}{7!} =\frac{7!}{(7-7)!} =\frac{7!}{0!} =\frac{7!}{1} =7!.\\5)A^{3} _{8} =\frac{8!}{(8-3)!} =\frac{8!}{5!} =\frac{5!*6*7*8}{5!} =6*7*8=336.\\6)A^{4} _{8} =\frac{8!}{(8-4)!} =\frac{8!}{4!} =\frac{4!*5*6*7*8}{4!} =5*6*7*8=1680.}$

$7)A^{2} _{10} =\frac{10!}{(10-2)!} =\frac{10!}{8!} =\frac{8!*9*10}{8!} =9*10=90.\\8)A^{4} _{10} =\frac{10!}{(10-4)!} =\frac{10!}{6!} =\frac{6!*7*8*9*10}{6!} =7*8*9*10=5040.\\330.\\A^{6} _{9} =\frac{9!}{(9-6)!} =\frac{9!}{3!}=\frac{3!*4*5*6*7*8*9}{3!}=4*5*6*7*8*9= 60480.\\331.\\A^{4} _{6}=\frac{6!}{( 6-4)!} =\frac{6!}{2!} =\frac{2!*3*4*5*6}{2!} =3*4*5*6=360.\\332.\\A^{2}_{30} =\frac{30!}{(30-2)!}=\frac{30!}{28!}=\frac{28!*29*30}{28!} } =29*30=870.$