Let the number of elements in sets $$A$$ and $$B$$ be five and two respectively. Then the number of subsets of $$A \times B$$ each having at least 3 and at most 6 elements is

To solve this, we use the property of Cartesian products and combinations.

1. Find the total elements in $$A \times B$$

The number of elements in the Cartesian product $$A \times B$$ is:

$$n(A \times B) = n(A) \times n(B) = 5 \times 2 = 10$$

2. Set up the Combination Sum

We need the number of subsets having "at least 3 and at most 6" elements. This is the sum of choosing 3, 4, 5, and 6 elements out of 10:

$$\text{Number of subsets} = ^{10}C_3 + ^{10}C_4 + ^{10}C_5 + ^{10}C_6$$

3. Calculate each term

• $$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$
• $$^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$
• $$^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$$
• $$^{10}C_6 = ^{10}C_4 = 210$$

4. Final Result

$$\text{Sum} = 120 + 210 + 252 + 210 = \mathbf{792}$$