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:
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:
Show 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: $\frac{n_A \cdot s_A}{k_A} = \frac{n_B \cdot s_B}{k_B}$.
Updated On: May 28, 2026
- 12
- 24
- 6
- 9
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The Correct Option is D
Solution and Explanation
Concept:
Counting problems involving set intersections and unions are solved by analyzing the total number of element slots filled across the groups. If we sum the sizes of all individual sets, each unique element is counted multiple times based on how many sets it belongs to.
Step 1: Calculate total slot counts via set population properties.
Let the total number of unique elements inside the combined union set $S$ be defined as $N = |S|$.
• We are given 6 sets of type $A$, where each set contains exactly 4 elements. Summing their individual sizes gives: \[ \sum_{i=1}^{6} |A_i| = 6 \times 4 = 24 \text{ slots} \]
• We are given $n$ sets of type $B$, where each set contains exactly 2 elements. Summing their individual sizes gives: \[ \sum_{j=1}^{n} |B_j| = n \times 2 = 2n \text{ slots} \]
Step 2: Relate the slot totals to the unique elements of set S.
The problem states how many times each element appears in each family of sets:
• Each unique element belongs to exactly 4 different $A$ sets. This means the total number of slots in family $A$ must equal $4N$: \[ 4N = 24 \quad \Rightarrow \quad N = 6 \] This confirms there are exactly 6 unique elements inside set $S$.
• Each unique element belongs to exactly 3 different $B$ sets. This means the total number of slots in family $B$ must equal $3N$: \[ 3N = 2n \]
Step 3: Solve for the unknown index number $n$.
Substitute our value of $N = 6$ into the balance equation: \[ 3(6) = 2n \quad \Rightarrow \quad 18 = 2n \quad \Rightarrow \quad n = 9 \] This matches option (D) perfectly.
Let the total number of unique elements inside the combined union set $S$ be defined as $N = |S|$.
• We are given 6 sets of type $A$, where each set contains exactly 4 elements. Summing their individual sizes gives: \[ \sum_{i=1}^{6} |A_i| = 6 \times 4 = 24 \text{ slots} \]
• We are given $n$ sets of type $B$, where each set contains exactly 2 elements. Summing their individual sizes gives: \[ \sum_{j=1}^{n} |B_j| = n \times 2 = 2n \text{ slots} \]
Step 2: Relate the slot totals to the unique elements of set S.
The problem states how many times each element appears in each family of sets:
• Each unique element belongs to exactly 4 different $A$ sets. This means the total number of slots in family $A$ must equal $4N$: \[ 4N = 24 \quad \Rightarrow \quad N = 6 \] This confirms there are exactly 6 unique elements inside set $S$.
• Each unique element belongs to exactly 3 different $B$ sets. This means the total number of slots in family $B$ must equal $3N$: \[ 3N = 2n \]
Step 3: Solve for the unknown index number $n$.
Substitute our value of $N = 6$ into the balance equation: \[ 3(6) = 2n \quad \Rightarrow \quad 18 = 2n \quad \Rightarrow \quad n = 9 \] This matches option (D) perfectly.
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