Define the set-theoretic structure that mathematically justifies combination calculations across disjoint sets | Step-by-Step Solution

What We're Solving:

We need to establish the rigorous set-theoretic foundation that justifies how we calculate combinations when selecting from two separate, disjoint sets. Essentially, we're building the mathematical "why" behind the formula ⁿCₖ₁ × ᵐCₖ₂.

The Approach:

We're building a mathematical house! We'll start with the foundation (basic set theory), then construct the walls (define combinations set-theoretically), and finally put on the roof (show why multiplication works for disjoint sets). This isn't just about getting an answer - we're proving why the mathematical structure behaves the way it does.

Step-by-Step Solution:

Step 1: Establish Our Foundation

• Start with two finite sets A and B where |A| = n, |B| = m, and A ∩ B = ∅ (disjoint)
• Define what we mean by "selecting k₁ elements from A and k₂ elements from B"

Step 2: Define Combinations Set-Theoretically

• For set A: C(A, k₁) = {S ⊆ A : |S| = k₁} (the collection of all k₁-element subsets)
• For set B: C(B, k₂) = {T ⊆ B : |T| = k₂} (the collection of all k₂-element subsets)
• Note that |C(A, k₁)| = ⁿCₖ₁ and |C(B, k₂)| = ᵐCₖ₂

Step 3: Construct the Cartesian Product Structure

• Define the combined selection space: C(A, k₁) × C(B, k₂)
• Each element is an ordered pair (S, T) where S ∈ C(A, k₁) and T ∈ C(B, k₂)
• This represents all possible ways to simultaneously choose k₁ from A and k₂ from B

Step 4: Apply the Fundamental Counting Principle

• Since A and B are disjoint, the selections are independent
• By the cardinality of Cartesian products: |C(A, k₁) × C(B, k₂)| = |C(A, k₁)| × |C(B, k₂)|
• Therefore: |C(A, k₁) × C(B, k₂)| = ⁿCₖ₁ × ᵐCₖ₂

Step 5: Verify Independence Property

• Show that the bijection f: C(A, k₁) × C(B, k₂) → P(A∪B) defined by f((S,T)) = S∪T preserves the structure
• Since A ∩ B = ∅, each union S∪T is uniquely determined by the pair (S,T)

The Answer:

The set-theoretic structure is: C(A, k₁) × C(B, k₂) = {(S, T) : S ⊆ A, T ⊆ B, |S| = k₁, |T| = k₂}

This Cartesian product structure has cardinality ⁿCₖ₁ × ᵐCₖ₂, which mathematically justifies why we multiply combinations when selecting from disjoint sets. The disjointness ensures independence, making the Cartesian product the natural mathematical framework.

Memory Tip:

Think "Disjoint sets = Independent choices = Multiply the counts!" The key insight is that disjointness allows us to use the Cartesian product structure, and the cardinality of a Cartesian product is always the product of the individual cardinalities. It's like choosing an outfit from separate closets - your shirt choice doesn't affect your pants options!

Remember, you're not just calculating - you're proving why the calculation works! 🌟