Q92 / 155 · WBJEE 2026 · Mathematics
Let $A_1, A_2, \ldots, A_6$ are six sets, each with four elements and $B_1, B_2, \ldots, B_n$ are $n$ sets, each with two elements. Let $S = A_1 \cup A_2 \cup \ldots \cup A_6 = B_1 \cup B_2 \cup \ldots \cup B_n$.
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
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$6$
- Correct answerD$9$
Solution
Let the number of elements in the set $S$ be $N$.
The sum of the cardinalities of the sets $A_1, A_2, \ldots, A_6$ is $6 \times 4 = 24$.
Since each element of $S$ belongs to exactly four of the sets $A_i$, the sum of the cardinalities is also $4N$.
Thus, $4N = 24 \Rightarrow N = 6$.
Similarly, the sum of the cardinalities of the sets $B_1, B_2, \ldots, B_n$ is $n \times 2 = 2n$.
Since each element of $S$ belongs to exactly three of the sets $B_j$, the sum of the cardinalities is also $3N$.
Thus, $3N = 2n$.
Substituting $N = 6$, we get $3 \times 6 = 2n \Rightarrow 18 = 2n \Rightarrow n = 9$.
Answer: $9$
The sum of the cardinalities of the sets $A_1, A_2, \ldots, A_6$ is $6 \times 4 = 24$.
Since each element of $S$ belongs to exactly four of the sets $A_i$, the sum of the cardinalities is also $4N$.
Thus, $4N = 24 \Rightarrow N = 6$.
Similarly, the sum of the cardinalities of the sets $B_1, B_2, \ldots, B_n$ is $n \times 2 = 2n$.
Since each element of $S$ belongs to exactly three of the sets $B_j$, the sum of the cardinalities is also $3N$.
Thus, $3N = 2n$.
Substituting $N = 6$, we get $3 \times 6 = 2n \Rightarrow 18 = 2n \Rightarrow n = 9$.
Answer: $9$
Hence, the correct answer is option: 'D'.