Problem 27 - A Complete Resource Book in Mathematics for JEE Main 2017

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 (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

Option (A) \( \frac{12!}{3!(4!)^3} \) is the correct answer.

Step 1: Calculate Total Arrangements of S

First, determine the total number of ways to arrange the elements of the set \( S = \{1, 2, 3, \ldots, 12\} \). There are \( 12! \) different ways to arrange these 12 elements.

Step 2: Split Arrangements into Groups

We want to partition the set \( S \) into three groups \( A, B, C \) such that each group has an equal size of 4 (since \( 12/3 = 4 \)). The order of groups does not matter, so we must divide by the number of arrangements of these groups.

Step 3: Accounting for Group Order

Since the order of the groups \( A, B, C \) does not matter, there are \( 3! \) different ways to arrange these groups. We divide \( 12! \) by \( 3! \) to account for these permutations.

Step 4: Calculate Individual Group Arrangements

Within each group of 4 elements, the arrangements do not matter. Therefore, we must divide by \( (4!)^3 \), as each group of 4 elements can be arranged in \( 4! \) different ways. This applies to the three groups \( A, B, \) and \( C \).

Step 5: Combine Steps to Determine Final Formula

The total number of ways to partition the set \( S \) is given by combining all these considerations:\[\frac{12!}{3! \times (4!)^3 }\]

Step 6: Choose Correct Answer

By comparing with the options given, the correct choice is:\[\text{(A) } \frac{12!}{3!(4!)^3}\]

Key Concepts

Set Partitioning

Set partitioning involves dividing a set into disjoint subsets such that the union of all subsets equals the original set. Each subset in a partition is called a "part." In our example, the set \( S = \{1, 2, 3, \ldots, 12\} \) is split into three subsets: \( A, B, \) and \( C \). Each of these parts needs to be of equal size, which in this case is 4, because we want each part to contain an equal number of elements such that \( |A| = |B| = |C| = 4 \).

Key conditions for set partitioning include:

• The union of all subsets is the entire original set.
• The intersection of any two distinct subsets is an empty set.

This means no element can be in more than one subset, making sure that the partition is complete and non-overlapping.

Arrangements

Arrangements refer to the different ways in which a set of elements can be ordered or organized. When arranging elements in a set, we consider the sequence in which the elements appear, which can change the perceived structure.

For the 12 elements in set \( S \), there are \( 12! \) (read as "12 factorial") ways to arrange these elements. However, when we partition them into groups where order does not matter, we need to adjust for redundancy. Hence, we must account for permutations within each subgroup and the order of the groups themselves.

So, if we have three groups, we consider \( 3! \) different possible orders. For each group of 4 elements, there are \( 4! \) ways to arrange them. This process helps eliminate redundant arrangements.

Factorial

The factorial of a number \( n \), denoted as \( n! \), is the product of all positive integers up to \( n \).
For example:

• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)

Factorials are crucial in combinatorics, as they help compute the number of possible arrangements or permutations of a set.

In our partitioning problem, \( 12! \) represents the total arrangements of all elements in set \( S \). We use factorials also to identify permutations within groups and between the groups themselves, such as \( 3! \) for the arrangement of groups \( A, B, C \), and \( (4!)^3 \) for the permutations within each group.
Understanding factorials and their properties is essential for solving problems that involve counting and arranging.

Grouping Elements

Grouping elements is a method of organizing elements into smaller collections that may adhere to certain criteria. In our scenario, we are grouping 12 elements into three separate parts, with each part containing 4 elements.

The goal of grouping here is to sample without replacement, ensuring that each element in set \( S \) only appears in one of the groups. The process involves:

• Selecting the first 4 elements from the 12 available. There are \( \binom{12}{4} \) ways to make this selection, which simplifies during solution steps to account for overlapping groups.
• Choosing the next set of 4 from the remaining 8 elements, \( \binom{8}{4} \), followed by the final group which is determined by default.

Correctly accounting for the methodology of arranging and grouping ensures that we accurately determine the number of ways such partitions can occur, which in our problem amounts to properly dividing \( 12! \) by \( 3! \times (4!)^3 \), giving us the answer as option (A).
This way, grouping keeps the subsets balanced and unique.