№322 Вычислите:

Решение:

1. 7! - 5!
   7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040
   5! = 5 * 4 * 3 * 2 * 1 = 120
   7! - 5! = 5040 - 120 = 4920
2. \[ \frac{24!}{20! \cdot 4!} \]
   \[ \frac{24!}{20! \cdot 4!} = \frac{24 \cdot 23 \cdot 22 \cdot 21 \cdot 20!}{20! \cdot (4 \cdot 3 \cdot 2 \cdot 1)} = \frac{24 \cdot 23 \cdot 22 \cdot 21}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{24}{4 × 3 × 2 × 1} × 23 × 22 × 21 = 1 × 23 × 22 × 21 = 10626 \]
3. \[ A_{5}^{2} = P_{5} \]
   Примечание: В задании, похоже, ошибка. Вероятно, имелось в виду \(A_{5}^{2} \) или \(P_{5}\). Будем считать, что \(A_{5}^{2}\).
   \[ A_{5}^{2} = \frac{5!}{(5-2)!} = \frac{5!}{3!} = \frac{5 × 4 × 3!}{3!} = 5 × 4 = 20 \]
   Если имелось в виду \(P_{5}\):
   \[ P_{5} = 5! = 5 × 4 × 3 × 2 × 1 = 120 \]
4. \[ C_{10}^{3} × P_{3} \]
   \[ C_{10}^{3} = \frac{10!}{3! × (10-3)!} = \frac{10!}{3! × 7!} = \frac{10 × 9 × 8 × 7!}{ (3 × 2 × 1) × 7!} = \frac{10 × 9 × 8}{3 × 2 × 1} = 10 × 3 × 4 = 120 \]
   \[ P_{3} = 3! = 3 × 2 × 1 = 6 \]
   Следовательно, \(C_{10}^{3} × P_{3} = 120 × 6 = 720 \]
5. \[ C_{11}^{2} + C_{11}^{3} \]
   \[ C_{11}^{2} = \frac{11!}{2! × (11-2)!} = \frac{11!}{2! × 9!} = \frac{11 × 10 × 9!}{ (2 × 1) × 9!} = \frac{11 × 10}{2} = 55 \]
   \[ C_{11}^{3} = \frac{11!}{3! × (11-3)!} = \frac{11!}{3! × 8!} = \frac{11 × 10 × 9 × 8!}{ (3 × 2 × 1) × 8!} = \frac{11 × 10 × 9}{6} = 11 × 5 × 3 = 165 \]
   Следовательно, \(C_{11}^{2} + C_{11}^{3} = 55 + 165 = 220 \]
6. \[ A_{7}^{3} + A_{7}^{4} \]
   \[ A_{7}^{3} = \frac{7!}{(7-3)!} = \frac{7!}{4!} = \frac{7 × 6 × 5 × 4!}{4!} = 7 × 6 × 5 = 210 \]
   \[ A_{7}^{4} = \frac{7!}{(7-4)!} = \frac{7!}{3!} = \frac{7 × 6 × 5 × 4 × 3!}{3!} = 7 × 6 × 5 × 4 = 840 \]
   Следовательно, \(A_{7}^{3} + A_{7}^{4} = 210 + 840 = 1050 \]