1) AP10-n P8 n ≤ 9; 2) P12 A1P14-n n ≤ 13.

Ответ: 1) (10-n)(9-n); 2) \(\frac{12!}{(13-n)! \cdot (14-n)!}\)

1. \( \frac{A_9^n \cdot P_{10-n}}{P_8} = \frac{\frac{9!}{(9-n)!} \cdot (10-n)!}{8!} = \frac{9! \cdot (10-n)!}{8! \cdot (9-n)!} = \frac{9 \cdot 8! \cdot (10-n)(9-n)(8-n)!}{8! \cdot (9-n)(8-n)!} = \frac{9(10-n)(9-n)}{9-n} = (10-n)(9-n) \)
2. \( \frac{P_{12}}{A_{13}^n \cdot P_{14-n}} = \frac{12!}{\frac{13!}{(13-n)!} \cdot (14-n)!} = \frac{12! \cdot (13-n)!}{13! \cdot (14-n)!} = \frac{12! \cdot (13-n)!}{13 \cdot 12! \cdot (14-n)!} = \frac{(13-n)!}{13 \cdot (14-n)!} = \frac{(13-n)!}{13 \cdot (14-n)(13-n)!} = \frac{1}{13 \cdot (14-n)} \)

Ответ: 1) (10-n)(9-n); 2) \(\frac{12!}{(13-n)! \cdot (14-n)!}\)