A · B = (4)(1) + (3)(2) = 4 + 6 = 10.

A · B = (1)(2) + (2)(3) + (3)(4) = 2 + 6 + 12 = 20.

A · B = (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32.

A · B = (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32

A · B = (1)(3) + (2)(4) = 3 + 8 = 11.

The number of ways to choose 4 students from 10 is 10C4 = 210.

The number of ways to choose 10 items from 15 is given by 15C10 = 15C5 = 3003.

The number of ways is 5C2 * 5C2 = 10 * 10 = 100.

The number of ways is 6C2 * 4C3 = 15 * 4 = 60.

The number of ways is 6C2 * 4C3 = 15 * 4 = 60.

The number of ways is 6C2 * 8C3 = 15 * 56 = 840.

The number of ways to select 2 men from 5 is 5C2 and 3 women from 6 is 6C3. Total = 5C2 * 6C3 = 10 * 20 = 200.

The total arrangements are 6! / (3! * 2! * 1!) = 60.

Treat the 4 men as one unit. Thus, we have 4 units (1 man unit + 3 women) to arrange: 4! * 4! = 24 * 24 = 576.

The number of ways to choose 4 students from 10 is given by 10C4 = 210.

The number of arrangements of 5 different flags is 5! = 120.

The number of arrangements around a circular table is (n-1)! = 5! = 120.

The number of arrangements of 6 people at a round table is (6-1)! = 5! = 120.

The number of ways to choose 2 fruits from 5 is given by 5C2 = 10.

Using the identity a^2 + b^2 = (a + b)^2 - 2ab, we have 70 = 12^2 - 2ab. Thus, 70 = 144 - 2ab, leading to ab = 37.

a² + b² = (a + b)² - 2ab = 5² - 2*6 = 25 - 12 = 13.

a³ + b³ = 1³ + 2³ = 1 + 8 = 9.

a³ + b³ = 1³ + 4³ = 1 + 64 = 65.

A · B = (1)(1) + (1)(1) + (1)(1) = 1 + 1 + 1 = 3.

A · B = (1)(1) + (1)(2) + (1)(3) = 1 + 2 + 3 = 6.

A · B = (1)(1) + (1)(2) + (1)(3) = 1 + 2 + 3 = 6.

A · B = (1)(2) + (1)(2) + (1)(2) = 2 + 2 + 2 = 6.

A · B = (1)(1) + (1)(-1) = 1 - 1 = 0.

A · B = (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32.

(a + b)² = (2 + 3)² = 25; a² + b² = 2² + 3² = 4 + 9 = 13; thus, 25 - 13 = 12.