6. Вычислить значения выражений: a) (8!-6!)/5! б) C^2_10 + C^3_10 в) A^2_5 * A^2_4 * A^3_2

Calculation Results

a) Calculation of the expression (8!-6!)/5!

• First, we calculate the factorials:
• 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
• 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
• 5! = 5 × 4 × 3 × 2 × 1 = 120
• Now, substitute these values into the expression:
• \[ \frac{8! - 6!}{5!} = \frac{40320 - 720}{120} \]
• Calculate the numerator:
• \[ 40320 - 720 = 39600 \]
• Now, divide by the denominator:
• \[ \frac{39600}{120} = 330 \]

b) Calculation of the expression C^2_10 + C^3_10

• We use the formula for combinations: C(n, k) = n! / (k!(n-k)!)
• Calculate C^2_10:
• \[ C_{10}^2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45 \]
• Calculate C^3_10:
• \[ C_{10}^3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
• Now, add the results:
• \[ C_{10}^2 + C_{10}^3 = 45 + 120 = 165 \]

c) Calculation of the expression A^2_5 * A^2_4 * A^3_2

• We use the formula for permutations: A(n, k) = n! / (n-k)!
• Calculate A^2_5:
• \[ A_5^2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20 \]
• Calculate A^2_4:
• \[ A_4^2 = \frac{4!}{(4-2)!} = \frac{4!}{2!} = 4 \times 3 = 12 \]
• Calculate A^3_2:
• \[ A_2^3 = \frac{2!}{(2-3)!} \text{ This is not possible as k > n} \]
• Since A^3_2 is not mathematically possible (you cannot choose 3 items from a set of 2), the entire expression is undefined or considered 0 in some contexts. Assuming this is a typo and it should be A^2_2 or similar, we will proceed by stating the impossibility of A^3_2.

Note: The term A³₂ is mathematically impossible because the number of items to choose (k=3) is greater than the total number of items available (n=2). Therefore, the entire expression in part (c) is not calculable as written.

Final Answers:

• a) 330
• b) 165
• c) Undefined (due to A³₂)