№329 Вычислить:

Решение:

1. \[ A_{8}^{3} + P_{3} \]
   \[ A_{8}^{3} = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 × 7 × 6 × 5!}{5!} = 8 × 7 × 6 = 336 \]
   \[ P_{3} = 3! = 3 × 2 × 1 = 6 \]
   Следовательно, \(A_{8}^{3} + P_{3} = 336 + 6 = 342 \]
2. \[ C_{5}^{2} - A_{6}^{1} + P_{6} \]
   \[ C_{5}^{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 × 4}{2 × 1} = 10 \]
   \[ A_{6}^{1} = \frac{6!}{(6-1)!} = \frac{6!}{5!} = 6 \]
   \[ P_{6} = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 \]
   Следовательно, \(C_{5}^{2} - A_{6}^{1} + P_{6} = 10 - 6 + 720 = 724 \]
3. \[ A_{11}^{2} + A_{10}^{4} \]
   \[ A_{11}^{2} = \frac{11!}{(11-2)!} = \frac{11!}{9!} = 11 × 10 = 110 \]
   \[ A_{10}^{4} = \frac{10!}{(10-4)!} = \frac{10!}{6!} = 10 × 9 × 8 × 7 = 5040 \]
   Следовательно, \(A_{11}^{2} + A_{10}^{4} = 110 + 5040 = 5150 \]
4. \[ C_{11}^{3} - C_{9}^{4} \]
   \[ C_{11}^{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} = \frac{11 × 10 × 9}{3 × 2 × 1} = 11 × 5 × 3 = 165 \]
   \[ C_{9}^{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 × 8 × 7 × 6}{4 × 3 × 2 × 1} = 9 × 2 × 7 = 126 \]
   Следовательно, \(C_{11}^{3} - C_{9}^{4} = 165 - 126 = 39 \]