Sets, Relations & Functions Question - None

Question

Let $n(X)$ denote the number of elements in $X$. If $A \cap B \cap C = \phi$, then find $n(A \cup B \cup C)$ in terms of $\sum n(A)$ and $\sum n(A \cap B)$. Also, given that $n(A \Delta B) = n(A) + n(B) - 2n(A \cap B)$, find $n(A \cup B \cup C)$ when $n(A \Delta B) = n(B \Delta C) = n(C \Delta A) = n(A) = n(B) = n(C)$. What is the value if the answer is 150?

Accepted Answer

Step 1: Formula for n(A ∪ B ∪ C) when A ∩ B ∩ C = ∅

By inclusion-exclusion principle: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

Since A ∩ B ∩ C = ∅: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A)

Step 2: Analyze the symmetric difference conditions

Given: n(A Δ B) = n(A) + n(B) - 2n(A ∩ B)

This represents elements in exactly one of A or B.

Let n(A) = n(B) = n(C) = k. We're told n(A Δ B) = n(B Δ C) = n(C Δ A) = k

Step 3: Set up equations from symmetric difference

From n(A Δ B) = k: k + k - 2n(A ∩ B) = k, so n(A ∩ B) = k/2

Similarly: n(B ∩ C) = k/2 and n(C ∩ A) = k/2

Step 4: Verify consistency with A ∩ B ∩ C = ∅

For each pairwise intersection to equal k/2 while their triple intersection is empty, each overlap region contains exactly k/2 elements belonging to exactly two sets (none in all three).

Step 5: Calculate n(A ∪ B ∪ C)

n(A ∪ B ∪ C) = k + k + k - k/2 - k/2 - k/2

= 3k - 3k/2

= 3k/2

Step 6: Find the specific value

If n(A ∪ B ∪ C) = 150, then 3k/2 = 150

Therefore: k = 100

This means n(A) = n(B) = n(C) = 100, and each pairwise intersection contains 50 elements.

Verification: n(A ∪ B ∪ C) = 100 + 100 + 100 - 50 - 50 - 50 = 300 - 150 = 150 ✓

∴ Answer: 150

Question:

Step-by-Step Solution

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