Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 466

\[1)\ {\overline{C}\ }_{4}^{6} = \frac{(6 + 4 - 1)!}{6!(4 - 1)!} = \frac{9!}{6! \bullet 3!} =\]

\[= \frac{9 \bullet 8 \bullet 7 \bullet 6!}{6! \bullet 3 \bullet 2} = 3 \bullet 4 \bullet 7 = 84;\]

\[2)\ {\overline{C}\ }_{6}^{4} = \frac{(4 + 6 - 1)!}{4!(6 - 1)!} = \frac{9!}{4! \bullet 5!} =\]

\[= \frac{9 \bullet 8 \bullet 7 \bullet 6 \bullet 5!}{4 \bullet 3 \bullet 2 \bullet 5!} = 3 \bullet 7 \bullet 6 = 126;\]

\[3)\ {\overline{C}\ }_{9}^{2} = \frac{(2 + 9 - 1)!}{2!(9 - 1)!} = \frac{10!}{2! \bullet 8!} =\]

\[= \frac{10 \bullet 9 \bullet 8!}{2 \bullet 8!} = 5 \bullet 9 = 45;\]

\[4)\ {\overline{C}\ }_{2}^{9} = \frac{(9 + 2 - 1)!}{9!(2 - 1)!} = \frac{10!}{9! \bullet 1!} =\]

\[= \frac{10 \bullet 9!}{9! \bullet 1} = 10.\]