№2. Вычислите число сочетаний 1) C³₃ = 2) C²₈ = 3) C⁵₈ = 4) C³₈ = 5) C¹²₁₂ =

• 1) $$C^3_3 = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} = \frac{3 \cdot 2 \cdot 1}{(3 \cdot 2 \cdot 1) \cdot 1} = 1$$
• 2) $$C^2_8 = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = \frac{8 \cdot 7 \cdot 6!}{(2 \cdot 1) \cdot 6!} = \frac{8 \cdot 7}{2} = \frac{56}{2} = 28$$
• 3) $$C^5_8 = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \cdot 7 \cdot 6 \cdot 5!}{5! \cdot 3 \cdot 2 \cdot 1} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = \frac{336}{6} = 56$$
• 4) $$C^3_8 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5!}{(3 \cdot 2 \cdot 1) \cdot 5!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = \frac{336}{6} = 56$$
• 5) $$C^{12}_{12} = \frac{12!}{12!(12-12)!} = \frac{12!}{12!0!} = \frac{12!}{12! \cdot 1} = 1$$

Ответ:

• 1) 1
• 2) 28
• 3) 56
• 4) 56
• 5) 1