Problem 75 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The set \(S=\\{1,2,3, \ldots \ldots, 12\\}\) 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 (a) \(\frac{12 !}{(4 !)^{3}}\) (b) \(\frac{12 !}{(4 !)^{4}}\) (c) \(\frac{12 !}{3 !(4 !)^{3}}\) (d) \(\frac{12 !}{3 !(4 !)^{4}}\)

Step-by-Step Solution

Answer

The number of ways is \( \frac{12!}{3!(4!)^3} \), which is option (c).

Step 1: Understand the Problem

We need to partition the set \( S = \{1, 2, 3, \ldots, 12\} \) into three subsets \( A, B, C \) such that each subset has 4 elements, and they are disjoint (i.e., \( A \cap B = B \cap C = A \cap C = \emptyset \)). Furthermore, \( A \cup B \cup C = S \). We need to determine how many such partitions exist.

Step 2: Calculate Total Arrangements

First, calculate the total way to arrange all 12 elements. This is the factorial of 12, which is: \[ 12! \].

Step 3: Divide by Internal Arrangements in Each Set

Each of the three sets \( A, B, \) and \( C \) have 4 elements. The number of ways to arrange each set internally is \( 4! \) for each. Thus, for three sets, we have: \[ (4!)^3 \].

Step 4: Account for Identical Sets

The sets \( A, B, \) and \( C \) are not distinguishable, meaning their arrangement doesn't matter. This introduces an additional division by \( 3! \) to account for the permutation of the sets themselves: \[ 3! \].

Step 5: Compute the Final Number of Partitions

Using the formula for the number of distinct partitions: \[ \frac{12!}{3! \cdot (4!)^3} \]. Expand and simplify each factorial to see why this is the correct expression.

Step 6: Determine the Correct Option

Upon simplifying \( \frac{12!}{3! \cdot (4!)^3} \), we verify it matches with option (c). Thus, the correct number of ways to partition the set is \( \frac{12!}{3!(4!)^3} \).

Key Concepts

Permutations in Combinatorics

Permutations are an essential concept in combinatorics, which deals with the arrangement of objects. When we talk about permutations, we focus on the different ways to order a set of items. These arrangements are significant in problems where the order of elements matters.

Let's consider the set we have in the original problem: \( S = \{1, 2, 3, \ldots, 12\} \). The total number of permutations of this set is represented by \( 12! \) (12 factorial). This number tells us how many different ways we can arrange all 12 elements sequentially. In general, for a set containing \( n \) distinct elements, the number of permutations is \( n! \).

Permutations are also used to define arrangements where some or all of the items are identical or indistinguishable. In the exercise, we face a situation where identical group arrangements (the sets \( A, B, \) and \( C \) having identical sizes) reduce the count of permutations by dividing by an additional factorial, \( 3! \), representing the indistinguishable groupings.

Understanding Factorials

Factorials are another cornerstone of combinatorics. A factorial, denoted with an exclamation mark \(!\), is the product of an integer and all the integers below it down to one.

For example, \( 4! \) (read as "four factorial") is \(4 \times 3 \times 2 \times 1 = 24\). Factorials are used to determine permutations, combinations, and a variety of other concepts in mathematics.

• \( n! \) gives the number of possible permutations of \( n \) distinct items.
• Factorials simplify the calculation of combinations, like dividing permutations by the factorials of group sizes as seen in the partitioning problem.

In our present problem, we used \( 12! \) to find all permutations of the set, and further \( (4!)^3 \) for arranging subsets \( A, B, \) and \( C \), each containing 4 elements. Factorials illuminate the symmetry and simplification in complex problems by efficiently tallying the number of outcomes.

Set Theory and Partitioning Sets

Set theory is a branch of mathematical logic that studies collections of objects known as sets. In this context, we often operate with terms like subsets, unions, and intersections. Particularly relevant here is the concept of partitioning a set.

When we partition a set, we divide it into non-overlapping and non-empty subsets so that their union reconstructs the original set. In the exercise, the set \( S \) is to be partitioned into three equal, disjoint subsets \( A, B, \) and \( C \). Each subset must be distinct, meaning they have no elements in common, satisfying \( A \cap B = B \cap C = A \cap C = \emptyset \).

Understanding how partitioning works involves recognizing that each element of the original set must end up in exactly one of the subsets. Furthermore, in combinatorics, correctly accounting for the distribution of elements across these subsets involves accounting for indistinguishable groups, which we deal with by dividing by factorials like \( 3! \) because sets \( A, B, \) and \( C \) can be swapped without affecting the overall arrangement.