Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1090

Авторы:
Год:2020-2021-2022-2023
Тип:учебник
Серия:Базовый и углубленный уровни

Задание 1090

\[\boxed{\mathbf{1090}\mathbf{.}}\]

\[\mathbf{Рекурентное\ свойство\ }\]

\[\mathbf{сочетаний}\mathbf{:}\]

\[C_{m}^{n} + C_{m}^{n + 1} = C_{m + 1}^{n + 1}\]

\[C_{m + 1}^{n + 1} - C_{m}^{n} = C_{m}^{n + 1}\ \ \]

\[C_{m + 1}^{n + 1} - C_{m}^{n + 1} = C_{m}^{n}\]

\[1)\ C_{13}^{10} + C_{13}^{11} = C_{14}^{11}\]

\[C_{14}^{11} = \frac{14!}{(14 - 11)! \bullet 11!} =\]

\[= \frac{14!}{3! \bullet 11!} = \frac{14 \bullet 13 \bullet 12 \bullet 11!}{3 \bullet 2 \bullet 11!} =\]

\[= 7 \bullet 13 \bullet 4 = 364.\]

\[2)\ C_{14}^{12} + C_{14}^{13} = C_{15}^{13}\]

\[C_{15}^{13} = \frac{15!}{(15 - 13)! \bullet 13!} =\]

\[= \frac{15!}{2! \bullet 13!} = \frac{15 \bullet 14 \bullet 13!}{2 \bullet 13!} =\]

\[= 15 \bullet 7 = 105.\]

\[3)\ C_{19}^{4} - C_{18}^{4} = C_{18}^{3}\]

\[C_{18}^{3} = \frac{18!}{(18 - 3)! \bullet 3!} = \frac{18!}{15! \bullet 3!} =\]

\[= \frac{18 \bullet 17 \bullet 16 \bullet 15!}{15! \bullet 3 \bullet 2} = 6 \bullet 17 \bullet 8 =\]

\[= 816.\]

\[4)\ C_{21}^{3} - C_{20}^{3} = C_{20}^{2}\]

\[C_{20}^{2} = \frac{20!}{(20 - 2)! \bullet 2!} = \frac{20!}{18! \bullet 2!} =\]

\[= \frac{20 \bullet 19 \bullet 18!}{18! \bullet 2} = 10 \bullet 19 = 190.\]

\[5)\ C_{61}^{3} - C_{60}^{2} = C_{60}^{3}\]

\[C_{60}^{3} = \frac{60!}{(60 - 3)! \bullet 3!} = \frac{60!}{57! \bullet 3!} =\]

\[= \frac{60 \bullet 59 \bullet 58 \bullet 57!}{57! \bullet 3 \bullet 2} =\]

\[= 20 \bullet 59 \bullet 29 = 34\ 220.\]

\[6)\ C_{71}^{3} - C_{70}^{2} = C_{70}^{3}\]

\[C_{70}^{3} = \frac{70!}{(70 - 3)! \bullet 3!} = \frac{70!}{67! \bullet 3!} =\]

\[= \frac{70 \bullet 69 \bullet 68 \bullet 67!}{67! \bullet 3 \bullet 2} =\]

\[= 35 \bullet 23 \bullet 68 = 54\ 740.\]

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