3. Вычислите: 1) $$\frac{6P_{11} - P_{10}}{13P_9}$$; 2) $$\frac{C_7^4}{A_6^3}$$

1) $$\frac{6P_{11} - P_{10}}{13P_9}$$

$$P_{11} = 11!$$
$$P_{10} = 10!$$
$$P_9 = 9!$$

$$\frac{6 \cdot 11! - 10!}{13 \cdot 9!} = \frac{6 \cdot 11 \cdot 10! - 10!}{13 \cdot 9!} = \frac{10!(6 \cdot 11 - 1)}{13 \cdot 9!} = \frac{10! \cdot 65}{13 \cdot 9!} = \frac{10 \cdot 9! \cdot 65}{13 \cdot 9!} = \frac{10 \cdot 65}{13} = 10 \cdot 5 = 50$$

Ответ: 50

2) $$\frac{C_7^4}{A_6^3}$$

$$C_7^4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \cdot 6 \cdot 5 \cdot 4!}{4! \cdot 3 \cdot 2 \cdot 1} = \frac{7 \cdot 6 \cdot 5}{6} = 35$$

$$A_6^3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{6 \cdot 5 \cdot 4 \cdot 3!}{3!} = 6 \cdot 5 \cdot 4 = 120$$

$$\frac{C_7^4}{A_6^3} = \frac{35}{120} = \frac{7}{24}$$

Ответ: 7/24