Задание 3. Вычислить значения выражений.

Решение:

Используем формулы бинома Ньютона и сочетаний для вычисления значений.

• 1. P₇; A³₁₀; C⁴₈: \( P_7 = 7! = 5040 \). \( A_7^3 = \frac{7!}{(7-3)!} = \frac{5040}{24} = 210 \). \( C_8^4 = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \).
• 2. P₈; A⁴₉ + C²₇: \( P_8 = 8! = 40320 \). \( A_9^4 = \frac{9!}{(9-4)!} = \frac{362880}{120} = 3024 \). \( C_7^2 = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \). \( 3024 + 21 = 3045 \).
• 3. P₉; A³₁₀ + C⁴₈: \( P_9 = 9! = 362880 \). \( A_{10}^3 = \frac{10!}{(10-3)!} = \frac{3628800}{5040} = 720 \). \( C_8^4 = \frac{8!}{4!(8-4)!} = 70 \). \( 720 + 70 = 790 \).
• 4. A³₁₀ + C⁴₈: \( A_{10}^3 = 720 \). \( C_8^4 = 70 \). \( 720 + 70 = 790 \).
• 5. P₃ - A²₇ + C⁵₇: \( P_3 = 3! = 6 \). \( A_7^2 = \frac{7!}{(7-2)!} = \frac{5040}{120} = 42 \). \( C_7^5 = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21 \). \( 6 - 42 + 21 = -15 \).
• 6. P₄ + A²₇ - C⁵₈: \( P_4 = 4! = 24 \). \( A_7^2 = 42 \). \( C_8^5 = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \). \( 24 + 42 - 56 = 10 \).
• 7. A²₇ + C⁴₇: \( A_7^2 = 42 \). \( C_7^4 = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \). \( 42 + 35 = 77 \).
• 8. A³₇ + C⁴₉: \( A_7^3 = \frac{7!}{(7-3)!} = \frac{5040}{24} = 210 \). \( C_9^4 = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \). \( 210 + 126 = 336 \).
• 9. P₅ - A²₇ + C³₆: \( P_5 = 5! = 120 \). \( A_7^2 = 42 \). \( C_6^3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \). \( 120 - 42 + 20 = 98 \).
• 10. P₇ + A³₈ + C⁴₇: \( P_7 = 5040 \). \( A_8^3 = \frac{8!}{(8-3)!} = \frac{40320}{120} = 336 \). \( C_7^4 = 35 \). \( 5040 + 336 + 35 = 5411 \).
• 11. P₈ + A⁴₈ + C²₆: \( P_8 = 40320 \). \( A_8^4 = \frac{8!}{(8-4)!} = \frac{40320}{24} = 1680 \). \( C_6^2 = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \). \( 40320 + 1680 + 15 = 42015 \).
• 12. P₄ + A³₉ + C⁴₉: \( P_4 = 24 \). \( A_9^3 = \frac{9!}{(9-3)!} = \frac{362880}{720} = 504 \). \( C_9^4 = 126 \). \( 24 + 504 + 126 = 654 \).
• 13. A³₁₀ + C⁴₈: \( A_{10}^3 = 720 \). \( C_8^4 = 70 \). \( 720 + 70 = 790 \).
• 14. P₇ - A⁴₁₀ + C⁵₉: \( P_7 = 5040 \). \( A_{10}^4 = \frac{10!}{(10-4)!} = \frac{3628800}{720} = 5040 \). \( C_9^5 = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \). \( 5040 - 5040 + 126 = 126 \).
• 15. P₆ + A²₉ - C⁵₉: \( P_6 = 6! = 720 \). \( A_9^2 = \frac{9!}{(9-2)!} = \frac{362880}{5040} = 72 \). \( C_9^5 = 126 \). \( 720 + 72 - 126 = 666 \).
• 16. A²₈ + C⁷₇: \( A_8^2 = \frac{8!}{(8-2)!} = \frac{40320}{720} = 56 \). \( C_7^7 = 1 \). \( 56 + 1 = 57 \).
• 17. A³₇ + C⁴₉: \( A_7^3 = 210 \). \( C_9^4 = 126 \). \( 210 + 126 = 336 \).
• 18. P₅ - A²₇ + C⁶₆: \( P_5 = 120 \). \( A_7^2 = 42 \). \( C_6^6 = 1 \). \( 120 - 42 + 1 = 79 \).
• 19. P₇ + A³₁₀ + C⁴₈: \( P_7 = 5040 \). \( A_{10}^3 = 720 \). \( C_8^4 = 70 \). \( 5040 + 720 + 70 = 5830 \).
• 20. P₈ + A⁴₉ + C²₇: \( P_8 = 40320 \). \( A_9^4 = 3024 \). \( C_7^2 = 21 \). \( 40320 + 3024 + 21 = 43365 \).
• 21. P₉ + A³₁₀ + C⁴₈: \( P_9 = 362880 \). \( A_{10}^3 = 720 \). \( C_8^4 = 70 \). \( 362880 + 720 + 70 = 363670 \).
• 22. A³₆ + C⁵₅: \( A_6^3 = \frac{6!}{(6-3)!} = \frac{720}{6} = 120 \). \( C_5^5 = 1 \). \( 120 + 1 = 121 \).
• 23. P₃ - A⁴₁₀ + C⁸₁₀: \( P_3 = 6 \). \( A_{10}^4 = 5040 \). \( C_{10}^8 = \frac{10!}{8!(10-8)!} = \frac{10 \times 9}{2 \times 1} = 45 \). \( 6 - 5040 + 45 = -4989 \).
• 24. P₆ + A³₇ - C⁶₈: \( P_6 = 720 \). \( A_7^3 = 210 \). \( C_8^6 = \frac{8!}{6!(8-6)!} = \frac{8 \times 7}{2 \times 1} = 28 \). \( 720 + 210 - 28 = 902 \).
• 25. P₅ - A²₇ + C⁶₆: \( P_5 = 120 \). \( A_7^2 = 42 \). \( C_6^6 = 1 \). \( 120 - 42 + 1 = 79 \).

Ответ: приведены вычисленные значения для каждого выражения.