Understanding the Problem: Subsets of Cartesian Product

The question asks us to find the number of subsets of the Cartesian product of two sets, A and B, which have specific sizes. We are given that set A has 4 elements and set B has 2 elements. The subsets we are interested in must contain at least three elements.

Calculating the Size of the Cartesian Product (A × B)

First, let's determine the total number of elements in the Cartesian product $A \times B$. The number of elements in the Cartesian product of two sets is the product of the number of elements in each set.

• Number of elements in set A, $|A| = 4$.
• Number of elements in set B, $|B| = 2$.
• The number of elements in the Cartesian product $A \times B$ is calculated as: $$ |A \times B| = |A| \times |B| $$ $$ |A \times B| = 4 \times 2 = 8 $$

So, the set $A \times B$ contains 8 elements.

Total Number of Subsets

A set with $n$ elements has a total of $2^n$ subsets. Since the set $A \times B$ has 8 elements, the total number of possible subsets is:

$$ \text{Total Subsets} = 2^{|A \times B|} = 2^8 $$

Calculating $2^8$:

$$ 2^8 = 256 $$

There are 256 possible subsets in total for the set $A \times B$. This includes subsets of all possible sizes, from 0 elements up to 8 elements.

Calculating Subsets with Fewer Than Three Elements

The question asks for subsets with *at least* three elements. It's easier to calculate the number of subsets that have *fewer than* three elements (i.e., 0, 1, or 2 elements) and subtract this from the total number of subsets.

The number of subsets with exactly $k$ elements from a set of $n$ elements is given by the binomial coefficient $\binom{n}{k}$. Here, $n=8$.

• Number of subsets with 0 elements: $$ \binom{8}{0} = \frac{8!}{0!(8-0)!} = 1 $$
• Number of subsets with 1 element: $$ \binom{8}{1} = \frac{8!}{1!(8-1)!} = \frac{8}{1} = 8 $$
• Number of subsets with 2 elements: $$ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 $$

The total number of subsets with fewer than three elements is the sum of these counts:

$$ \text{Subsets with < 3 elements} = \binom{8}{0} + \binom{8}{1} + \binom{8}{2} $$ $$ \text{Subsets with < 3 elements} = 1 + 8 + 28 = 37 $$

Calculating Subsets with At Least Three Elements

To find the number of subsets with at least three elements, we subtract the count of subsets with fewer than three elements from the total number of subsets.

$$ \text{Subsets with } \ge 3 \text{ elements} = \text{Total Subsets} - (\text{Subsets with < 3 elements}) $$ $$ \text{Subsets with } \ge 3 \text{ elements} = 256 - 37 $$ $$ \text{Subsets with } \ge 3 \text{ elements} = 219 $$

Final Answer

Therefore, the number of subsets of the set $A \times B$, each having at least three elements, is 219.