Problem 55 - A Complete Resource Book in Mathematics for JEE Main

Question

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such arrangements is (A) less than 500 (B) at least 500 but less than 750 (C) at least 750 but less than 1000 (D) at least 1000

Step-by-Step Solution

Answer

(D) at least 1000

Step 1: Select the novels

We have 6 novels and need to select 4 of them. The number of ways to select 4 novels from 6 is given by the combination formula: \( \binom{6}{4} \). This calculates to \( \binom{6}{4} = 15 \).

Step 2: Select the dictionary

From the 3 dictionaries available, 1 needs to be selected. There are \( \binom{3}{1} = 3 \) ways to choose 1 dictionary from 3.

Step 3: Arrange novels and place the dictionary

First, arrange the 4 selected novels in a sequence, which can be done in \( 4! \) ways. The dictionary must occupy the middle position (the 3rd position in the sequence). Arranging 4 novels gives us \( 24 \) permutations.

Step 4: Calculate total arrangements

In total, the number of arrangements is given by the product of the selections and permutations calculated previously: \( \binom{6}{4} \times \binom{3}{1} \times 4! = 15 \times 3 \times 24 \). This equals 1080 arrangements.

Key Concepts

Permutation and Combination

In mathematics, **Permutations** and **Combinations** are tools to determine various ways to arrange or select items from a set. This concept often finds application when understanding how to choose and order books or any objects.

- **Combination** finds the number of ways to choose a subset of items from a larger set, where the order doesn't matter. In the exercise, we identified the total ways to select 4 novels from 6. This selection process where order is irrelevant is calculated using the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Here, \( n \) is the total number of items, and \( r \) is the number of items to choose. So, for choosing 4 novels from 6, it is \( \binom{6}{4} = 15 \) ways.
- **Permutation**, on the other hand, regards the arrangement of subsets where the order does matter. It wasn't directly highlighted in the step, but once novels are chosen, arranging them is critical, as shown in factorial calculations. Understanding how choices and ordering affect outcomes is vital.

Factorial Calculations

Factorials are fundamental in combinatorics for calculating permutations and combinations.

- **Factorial** of a non-negative integer \( n \), denoted by \( n! \), is the product of all positive integers less than or equal to \( n \). For instance, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). This operation helps compute how many different ways items can be arranged. For instance, organizing 4 novels is possible in \( 24 \) different sequences.
- In the given exercise, once the novels were selected, their sequence had to be set, which was computed using \( 4! \). Since the dictionary must always be in the middle, no factorial is needed for its placement; its position is fixed. Nevertheless, understanding and calculating factorials is essential to grasp permutations and combinations effectively.

Arrangement Problems in Mathematics

Arrangement problems involve finding methods to order different items according to specific rules or conditions. These problems require both understanding and application of permutation and combination principles.

For this exercise, books were to be arranged along a shelf with specific constraints:

• Four novels had to be chosen from a set, dictating use of the combination formula, \( \binom{6}{4} \).
• They also had to be arranged in a certain order, using the permutation principle where \( 4! \) ways was applied.
• The dictionary, selected from 3 options, must occupy the third position, giving importance to arrangement rules.

By binding selection with arrangement, we can solve complex problems methodically.
The total number of ways to accomplish this was a product of the selections and arrangements: \( 15 \times 3 \times 24 = 1080 \). This showcases how combination, arrangement, and the structure of permutations can solve practical arrangement problems.