Упр.737 ГДЗ Колягин Ткачёва 11 класс (Алгебра)
1) A11 3 — A10 2/A9 1;
2) A12 4 * A7 7/A11 9;
3) A6 3/P4 + A11 6/11P6;
4) (C11 7/10 — C5 2/10)P6/A6 4.
$$\frac{A_{11}^{3}-A_{10}^{2}}{A_{9}^{1}}=\frac{\frac{11!}{(11-3)!}-\frac{10!}{(10-2)!}}{\frac{9!}{(9-1)!}}=\frac{\frac{11!}{8!}-\frac{10!}{8!}}{\frac{9!}{8!}}=\frac{11!-10!}{9!}$$
$$\frac{11\cdot 10! — 10!}{9!}=\frac{10\cdot 10!}{9!}=\frac{10\cdot 10\cdot 9!}{9!}=100$$
$$\frac{A_{12}^{4}\cdot A_{7}^{7}}{A_{11}^{9}}=\frac{\frac{12!}{(12-4)!}\cdot \frac{7!}{(7-7)!}}{\frac{11!}{(11-9)!}}=\frac{\frac{12!}{8!}\cdot 7!}{\frac{11!}{2!}}$$
$$\frac{12\cdot 11\cdot 10\cdot 9\cdot 7! \cdot 2!}{8!\cdot 11!}=\frac{12}{8}\cdot 2=3$$
$$\frac{A_{6}^{3}}{P_{4}}+\frac{A_{11}^{6}}{11P_{6}}=\frac{\frac{6!}{(6-3)!}}{4!}+\frac{\frac{11!}{(11-6)!}}{11\cdot 6!}$$
$$=\frac{6!}{3!\cdot 4!}+\frac{11!}{5!\cdot 11\cdot 6!}=\frac{6\cdot 5\cdot 4!}{3!\cdot 4!}+\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6!}{5!\cdot 11\cdot 6!}$$
$$=5+42=47$$
$$\left(\frac{C_{11}^{7}}{10}-\frac{C_{5}^{2}}{10}\right)\frac{P_{6}}{A_{6}^{4}}=\left(\frac{\frac{11!}{7!\,4!}}{10}-\frac{\frac{5!}{2!\,3!}}{10}\right)\cdot \frac{6!}{\frac{6!}{2!}}$$
$$=\left(\frac{11\cdot 10\cdot 9\cdot 8}{4!\cdot 10}-\frac{5\cdot 4}{2\cdot 10}\right)\cdot 2=\left(\frac{11\cdot 9\cdot 8}{4\cdot 3\cdot 2}-1\right)\cdot 2$$
$$=(33-1)\cdot 2=64$$
Ответ
1) 100; 2) 3; 3) 47; 4) 64.