Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 440

\[1)\ \frac{A_{11}^{3} - A_{10}^{2}}{A_{9}^{1}} =\]

\[= \frac{\frac{11!}{(11 - 3)!} - \frac{10!}{(10 - 2)!}}{\frac{9!}{(9 - 1)!}} =\]

\[= \frac{\frac{11!}{8!} - \frac{10!}{8!}}{\frac{9!}{8!}} = \frac{11! - 10!}{9!} =\]

\[= \frac{11 \bullet 10! - 10!}{9!} = \frac{10 \bullet 10!}{9!} =\]

\[= \frac{10 \bullet 10 \bullet 9!}{9!} = 100.\]

\[2)\ \frac{A_{12}^{4} \bullet A_{7}^{7}}{A_{11}^{9}} = \frac{\frac{12!}{(12 - 4)!} \bullet P_{7}}{\frac{11!}{(11 - 9)!}} =\]

\[= \frac{\frac{12!}{8!} \bullet 7!}{\frac{11!}{2!}} = \frac{\frac{12!}{8 \bullet 7!} \bullet 7!}{\frac{11!}{2}} =\]

\[= \frac{12!}{8}\ :\frac{11!}{2} = \frac{12! \bullet 2}{8 \bullet 11!} =\]

\[= \frac{12 \bullet 11! \bullet 2}{8 \bullet 11!} = \frac{12}{4} = 3.\]

\[Ответ:\ \ 3.\]