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

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$$\mathbf{Р}\mathbf{е}\mathbf{к}\mathbf{у}\mathbf{р}\mathbf{е}\mathbf{н}\mathbf{т}\mathbf{н}\mathbf{о}\mathbf{е}\mathbf{с}\mathbf{в}\mathbf{о}\mathbf{й}\mathbf{с}\mathbf{т}\mathbf{в}\mathbf{о}$$

$$\mathbf{с}\mathbf{о}\mathbf{ч}\mathbf{е}\mathbf{т}\mathbf{а}\mathbf{н}\mathbf{и}\mathbf{й}\mathbf{:}$$

$$C_m^n+C_m^{n+1}=C_{m+1}^{n+1}$$

$$C_{m+1}^{n+1}-C_m^n=C_m^{n+1}$$

$$C_{m+1}^{n+1}-C_m^{n+1}=C_m^n$$

$$1)C_{13}^{10}+C_{13}^{11}=C_{14}^{11}$$

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

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

$$=7\bullet 13\bullet 4=364.$$

$$2)C_{14}^{12}+C_{14}^{13}=C_{15}^{13}$$

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

$$=\frac{15!}{2!\bullet 13!}=\frac{15\bullet 14\bullet 13!}{2\bullet 13!}=$$

$$=15\bullet 7=105.$$

$$3)C_{19}^4-C_{18}^4=C_{18}^3$$

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

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

$$=816.$$

$$4)C_{21}^3-C_{20}^3=C_{20}^2$$

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

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

$$5)C_{61}^3-C_{60}^2=C_{60}^3$$

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

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

$$=20\bullet 59\bullet 29=34220.$$

$$6)C_{71}^3-C_{70}^2=C_{70}^3$$

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

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

$$=35\bullet 23\bullet 68=54740.$$