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Question #41090

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be a set of 9 distinct elements. Let $P_1, P_2,$ and $P_3$ be three mutually disjoint 2-element subsets of $S$ . The elements of $S$ are to be partitioned into 3 unlabelled subsets of 3 elements each. Find the number of such partitions where at least one of the subsets $P_1, P_2, P_3$ is completely contained within one of the 3-element subsets of the partition.

Options

A.
186
B.
156
C.
210
D.
366

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Inclusion-Exclusion Principle|Restricted Selection (At least one common)|Division into Groups

Why This Question Is Tricky

Common Mistakes

This problem tests the Inclusion-Exclusion Principle combined with the distribution of distinct items into identical groups.

The primary trap lies in the calculation of $|E_1|$ and $|E_1 \cap E_2|$ .

Students often forget to divide by $2!$ when partitioning the remaining 6 items into two identical groups of 3 for $|E_1|$ , leading to a massive overcount (366).

Conversely, they might erroneously divide by $2!$ in $|E_1 \cap E_2|$ , failing to realize that the presence of distinct pairs $P_1$ and $P_2$ acts as unique labels, making the groups distinguishable (leading to 186).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Inclusion-Exclusion Principle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free