Problem 30 - Algebra for JEE ADVANCED

Question

Three boys of class \(\mathrm{X}\), four boys of class \(\mathrm{XI}\) and five boys of class XII sit in a row. The total number of ways in which these boys can sit so that all the boys of same class sit together is equal to a. \((3 !)^{2}(4 !)(5 !)\) b. \((3 !)(4 !)^{2}(5 !)\) c. \((3 !)(4 !)(5 !)\) d. (3!) \((4 !)(5 !)^{2}\)

Step-by-Step Solution

Answer

Correct option is (a): \((3!)^2(4!)(5!)\)

Step 1: Identify Groups

First, consider each class of boys as a single group (block) for seating purposes. Therefore, we have 3 groups: one with 3 boys from class X, one with 4 boys from class XI, and one with 5 boys from class XII.

Step 2: Arrange the Groups

We need to arrange these 3 groups. The number of ways to arrange 3 distinct groups is given by the factorial of 3, which is: \[3! = 6\]

Step 3: Arrange Boys Within Each Group

Within each group, boys can sit in any order. For the group of class X with 3 boys, there are:\[3! = 6\]ways. For the group of class XI with 4 boys, there are:\[4! = 24\]ways. For the group of class XII with 5 boys, there are:\[5! = 120\]ways.

Step 4: Calculate Total Arrangements

To find the total number of arrangements, multiply the number of ways to arrange the groups by the number of ways to arrange boys within each group:\[3! imes 3! imes 4! imes 5! = 6 imes 6 imes 24 imes 120\]

Step 5: Simplify the Expression

Simplify the calculated expression for the total number of arrangements:\[3!^2 imes 4! imes 5!\] Which involves calculating:\[6 imes 6 = 36\]Then:\[36 imes 24 = 864\]And finally:\[864 \times 120 = 103680\]

Step 6: Select the Correct Option

Compare your calculated expression \(3!^2 \times 4! \times 5!\) with the given options. The correct multiple-choice option is provided by this calculation, which matches option (a).

Key Concepts

Factorial

The concept of factorial is fundamental in permutations and combinations. The factorial of a number, represented as \( n! \), is the product of all positive integers from 1 to \( n \). For example, the factorial of 3 is \( 3! = 3 \times 2 \times 1 = 6 \). Factorials allow us to determine the number of ways to arrange \( n \) items uniquely. When calculating arrangements, the factorial function grows rapidly.
For instance, arranging 5 boys involves computing \( 5! = 120 \). This quick escalation in factorial values reflects the increasing complexity and vast number of possibilities when we deal with larger groups in permutation problems.

Group Theory

Group theory provides a mathematical framework for understanding symmetry and arrangement. In problems involving permutations and combinations, we use group theory to deal with the arrangements of elements where certain conditions hold. In our exercise, the boys from each class can be considered as a group. These groups need to maintain their structure while varying positions among themselves.
Considering each group as a single block or entity simplifies the problem by reducing the number of entities we must permute. In our case with three groups (class X, XI, XII), group theory helps us determine that we can rearrange them in \( 3! = 6 \) different ways. This initial simplification is crucial for facilitation of further calculations.

Arrangement Problems

Arrangement problems, like the one we are analyzing, involve finding all possible configurations under specific conditions. These problems often utilize factorial calculations to account for different seating orders.
The problem requires us to ensure that boys from the same class sit together, effectively transforming them into blocks. After establishing blocks for each class, we apply permutation techniques to arrange these blocks and then arrange the boys within each block.

• This involves calculating individual group arrangements \( 3! \), \( 4! \), and \( 5! \) for classes X, XI, and XII respectively.
• Then, multiplying these by the arrangement of blocks \( 3! \), capturing the total possible scenarios.

This methodical approach solves arrangement problems efficiently, ensuring all conditions are met while calculating the vast number of different possible orders.