<p>In how many different ways can a set \(A\) of \(3n\) elements be partitioned into 3 subsets of equal number of elements? (The subsets \(P, Q, R\) form a partition if \(P \cup Q \cup R = A\), \(P \cap R = \phi\), \(Q \cap R = \phi\), \(R \cap P = \phi\).)</p>
Step-by-Step Solution
Key Concept: When partitioning into indistinguishable groups, divide by the factorial of the number of groups to eliminate overcounting from different orderings of identical-sized subsets. The formula is C(3n,n)·C(2n,n)·C(n,n) divided by 3! since the three subsets are unlabeled.
<p><strong>Step 1:</strong> Recognize we need to partition 3n elements into 3 equal unlabeled subsets of n elements each.</p><p><strong>Step 2:</strong> If subsets were labeled (distinguishable), we'd choose: C(3n,n) ways to select first n elements, then C(2n,n) ways to select next n from remaining 2n, then C(n,n)=1 way for the last subset.</p><p><strong>Step 3:</strong> Calculate: C(3n,n)·C(2n,n) = $\dfrac{(3n)!}{n! \cdot (2n)!} \cdot \dfrac{(2n)!}{n! \cdot n!} = \dfrac{(3n)!}{(n!)^3}$</p><p><strong>Step 4:</strong> Since the three subsets are <strong>unlabeled/indistinguishable</strong>, we must divide by 3! to account for the overcounting from all possible orderings of the same three subsets.</p><p><strong>Step 5:</strong> Final answer: $\dfrac{(3n)!}{(n!)^3 \cdot 3!}$</p><p>∴ Answer: $\dfrac{3n!}{(n!)^3 \cdot 3!}$</p>
Correct Answer: \(\dfrac{3n!}{(n!)^3 \cdot 3!}\)