Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 435

\[1)\ A_{3}^{1} = \frac{3!}{(3 - 1)!} = \frac{3!}{2!} = \frac{3 \bullet 2!}{2!} = 3;\]

\[2)\ A_{3}^{2} = \frac{3!}{(3 - 2)!} = \frac{3!}{1!} = 3 \bullet 2 = 6;\]

\[3)\ A_{7}^{2} = \frac{7!}{(7 - 2)!} = \frac{7!}{5!} =\]

\[= \frac{7 \bullet 6 \bullet 5!}{5!} = 42;\]

\[4)\ A_{7}^{7} = P_{7} = 7! =\]

\[= 7 \bullet 6 \bullet 5 \bullet 4 \bullet 3 \bullet 2 = 5040;\]

\[5)\ A_{8}^{3} = \frac{8!}{(8 - 3)!} =\]

\[= \frac{8 \bullet 7 \bullet 6 \bullet 5!}{5!} = 336;\]

\[6)\ A_{8}^{4} = \frac{8!}{(8 - 4)!} =\]

\[= \frac{8 \bullet 7 \bullet 6 \bullet 5 \bullet 4!}{4!} = 1680;\]

\[7)\ A_{10}^{2} = \frac{10!}{(10 - 2)!} =\]

\[= \frac{10 \bullet 9 \bullet 8!}{8!} = 90;\]

\[8)\ A_{10}^{4} = \frac{10!}{(10 - 4)!} =\]

\[= \frac{10 \bullet 9 \bullet 8 \bullet 7 \bullet 6!}{6!} = 5040.\]