Problem 27 - A Complete Resource Book in Mathematics for JEE Main

Question

The set \(S=\\{1,2,3, \ldots, 12\) ) is to be partitioned into three sets \(A, B, C\) of equal size. Thus, \(A \cup B \cup C=S\), \(A \cap B=B \cap C=A \cap C=\phi .\) The number of ways to partition \(\mathrm{S}\) is \(\quad[\mathbf{2 0 0 7}]\) (A) \(\frac{12 !}{3 !(4 !)^{3}}\) (B) \(\frac{12 !}{3 !(3 !)^{4}}\) (C) \(\frac{12 !}{(4 !)^{3}}\) (D) \(\frac{12 !}{(3 !)^{4}}\)

Step-by-Step Solution

Answer

(A) \( \frac{12!}{3! (4!)^3} \)

Step 1: Understand the Problem

We need to partition the set \( S = \{1, 2, 3, \ldots, 12\} \) into three disjoint subsets \( A, B, \) and \( C \), each containing four elements, such that \( A \cup B \cup C = S \) and \( A \cap B = B \cap C = A \cap C = \emptyset \). We must calculate the total number of possible partitions.

Step 2: Calculate Total Arrangements

First, find the total number of ways to arrange the 12 elements. This is given by \( 12! \).

Step 3: Divide into Groups of 4

Since each subset \( A, B, C \) must contain exactly 4 elements, divide the arrangements by \( 4! \) for each subset to account for the indistinct order within each subset. Thus, divide by \( (4!)^3 \).

Step 4: Account for Exchangeable Subsets

Since the sets \( A, B, C \) can be exchanged internally, divide the arrangements by the number of ways to arrange the three sets, which is \( 3! \).

Step 5: Combine Calculations

Combine the calculations to find the number of ways to partition:\[\frac{12!}{3! \times (4!)^3}\]

Step 6: Identify the Correct Choice

Compare our result with the provided options. The result \( \frac{12!}{3! \times (4!)^3} \) matches option (A).

Key Concepts

Set Partitioning

Set partitioning is a fundamental concept in combinatorics, where we divide a set into distinct subsets. Each subset is disjoint, meaning they have no elements in common, and together they complete the original set.
For example, in our problem, we want to divide a set \( S = \{1, 2, 3, \ldots, 12\} \) into three subsets \( A, B, \) and \( C \). Each of these subsets needs to have exactly four elements.
Important points to consider in set partitioning:

• The subsets must be mutually exclusive, which means the intersection between any two of them is empty: \( A \cap B = B \cap C = A \cap C = \emptyset \).
• Every element in the original set must be in one of the subsets, ensuring \( A \cup B \cup C = S \).

Set partitioning helps in organizing data and understanding how combinations can be structured when specific conditions are applied.

Factorial Calculations

Factorial calculations are crucial in determining the number of ways to arrange or select items. The factorial of a number \( n \) is denoted as \( n! \) and is the product of all positive integers up to \( n \).
For our set \( S \), which has 12 elements, the total number of arrangements is \( 12! \). This represents every possible way to order the elements in the set.
When partitioning our set into smaller groups, we need to account for the internal arrangement of each subset. Each subset has 4 elements, so we calculate \( 4! \) for each of the three subsets to find the arrangement within them.

• \( 12! \) gives the total permutations of the whole set.
• \( (4!)^3 \) accounts for arrangements within each subset \( A, B, \) and \( C \).
• Dividing by \( 3! \) further accounts for the interchangeable positions of the subsets themselves.

This is why the solution involves dividing \( 12! \) by \( 3! \times (4!)^3 \), as this accounts for both internal arrangements within subsets and the external exchange of these subsets.

Disjoint Subsets

Disjoint subsets play a significant role in partitioning a set. Disjoint subsets are subsets that do not share any elements. In our context, \( A, B, \) and \( C \) are disjoint because \( A \cap B = B \cap C = A \cap C = \emptyset \).
In practice, disjoint subsets help ensure that every element is uniquely allocated to one and only one subset. This is crucial in scenarios where overlapping, or shared membership, would invalidate the criteria of the partition.
In our exercise:

• The disjoint nature of \( A, B, \) and \( C \) ensures that their union forms the entire set \( S \) without omissions or duplications.
• Each element from \( S \) is accounted for uniquely in either \( A, B, \) or \( C \).

This principle is an essential aspect of ensuring we adhere to the conditions given when partitioning sets, especially in structured data organization and problem-solving within combinatorics.