Understanding the Problem

We are given two sets: A has 4 elements and B has 2 elements. The Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B. The number of elements in A × B is |A| × |B| = 4 × 2 = 8. So, A × B has 8 elements.

We need to find the number of subsets of A × B that have at least 3 elements. Since A × B has 8 elements, the total number of subsets is 2⁸ = 256. However, we are only interested in subsets with size 3 or more.

Step-by-Step Solution

Step 1: Find the total number of subsets of A × B.

For a set with n elements, the number of subsets is 2ⁿ. Here, n = 8.

Total subsets = $2^8=256$

Step 2: Find the number of subsets with fewer than 3 elements (i.e., sizes 0, 1, and 2).

The number of subsets of size k from a set of n elements is given by the combination formula: $C(n,k)=\frac{n!}{k!(n-k)!}$

For n = 8:

• Subsets of size 0: $C(8,0)=1$
• Subsets of size 1: $C(8,1)=8$
• Subsets of size 2: $C(8,2)=\frac{8\times 7}{2\times 1}=28$

Total subsets with fewer than 3 elements = 1 + 8 + 28 = 37

Step 3: Subtract the number of subsets with fewer than 3 elements from the total number of subsets.

Number of subsets with at least 3 elements = Total subsets - Subsets with size < 3

= 256 - 37 = 219

Final Answer: 219

Related Topics

• Set Theory: Cartesian product, subsets, power set.
• Combinatorics: Combinations, binomial coefficients.

Key Formulae

• Number of elements in Cartesian product: |A × B| = |A| × |B|
• Number of subsets of a set with n elements: 2ⁿ
• Number of combinations (subsets of size k): $C(n,k)=\frac{n!}{k!(n-k)!}$