Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1105

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\[1)\ C_{12}^{10} + C_{12}^{11} = C_{13}^{11} =\]

\[= \frac{13!}{(13 - 11)! \bullet 11!} =\]

\[= \frac{13 \bullet 12 \bullet 11!}{2! \bullet 11!} = 13 \bullet 6 = 78;\]

\[2)\ C_{11}^{9} + C_{11}^{10} = C_{12}^{10} =\]

\[= \frac{12!}{(12 - 10)! \bullet 10!} =\]

\[= \frac{12 \bullet 11 \bullet 10!}{2! \bullet 10!} = 6 \bullet 11 = 66;\]

\[3)\ C_{8}^{6} + C_{8}^{7} + C_{9}^{8} = C_{9}^{7} + C_{9}^{8} =\]

\[= C_{10}^{8} = \frac{10!}{(10 - 8)! \bullet 8!} =\]

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

\[4)\ C_{9}^{5} + C_{9}^{6} + C_{10}^{7} = C_{10}^{6} + C_{10}^{7} =\]

\[= C_{11}^{7} = \frac{11!}{(11 - 7)! \bullet 7!} = \frac{11!}{4! \bullet 7!} =\]

\[= \frac{11 \bullet 10 \bullet 9 \bullet 8 \bullet 7!}{4 \bullet 3 \bullet 2 \bullet 7!} =\]

\[= 11 \bullet 10 \bullet 3 = 330.\]