Question

Let \(A_{1},A_{2},\dots,A_{6}\) be six sets, each with four elements and \(B_{1},B_{2},\dots,B_{n}\) be \(n\) sets, each with two elements. Let \[ S=A_{1}\cup A_{2}\cup\dots\cup A_{6}=B_{1}\cup B_{2}\cup\dots\cup B_{n}. \] Given that each element of \(S\) belongs to exactly four of the \(A\)'s and to exactly three of the \(B\)'s, then \(n\) is:

• A. 12
• B. 24
• C. 6
• D. 9

Hint:

The general shortcut formula for this type of set distribution problem is $N \cdot k_A = n_A \cdot s_A$ and $N \cdot k_B = n_B \cdot s_B$, where $s$ is the set size and $k$ is the membership frequency. Equating them gives the direct relation: $\frac{n_A \cdot s_A}{k_A} = \frac{n_B \cdot s_B}{k_B}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a problem based on the counting of elements in a union of sets with overlaps. The total count of elements across all sets (counting repeats) must equal the number of distinct elements multiplied by the frequency of each element's appearance.
Step 2: Key Formula or Approach:
1. Let \( |S| = m \).
2. Use the "double counting" principle: \( \sum |A_i| = m \times (\text{frequency in A}) \) and \( \sum |B_j| = m \times (\text{frequency in B}) \).
Step 3: Detailed Explanation:
Calculate total entries in the A-collection:
There are 6 sets, each with 4 elements. Total entries \( = 6 \times 4 = 24 \).
Since each distinct element of \( S \) belongs to exactly 4 sets in the A-collection:
\[ 4 \times |S| = 24 \implies |S| = 6 \]
Now, look at the B-collection:
There are \( n \) sets, each with 2 elements. Total entries \( = n \times 2 = 2n \).
Each distinct element of \( S \) (we found \( |S| = 6 \)) belongs to exactly 3 sets in the B-collection.
Therefore:
\[ 3 \times |S| = 2n \]
\[ 3 \times 6 = 2n \]
\[ 18 = 2n \implies n = 9 \]
Step 4: Final Answer:
The number of sets in the second collection is 9.