Problem 35 - A Complete Resource Book in Mathematics for JEE Main

Question

If all permutations of the letters of the word \(A G A I N\) are arranged as in dictionary, the forty ninth word is (A) NA \(\overrightarrow{A G I}\) (B) \(N A G A I\) (C) NAAIG (D) \(N A I A G\)

Step-by-Step Solution

Answer

The 49th word is C: NAAIG.

Step 1: Determine the Total Number of Permutations

First, calculate the total number of permutations of the letters in the word "AGAIN". The word 'AGAIN' consists of 5 letters with 'A' repeated twice. The formula for permutations of a word with repeated letters is \( \frac{n!}{p!} \), where \( n \) is the total number of letters and \( p \) is the factorial of the number of times each repeated letter appears. So, for "AGAIN": \( \frac{5!}{2!} = \frac{120}{2} = 60 \) permutations.

Step 2: Determine Number of Permutations Starting with Each Letter

Arrange the permutations as would appear in a dictionary. First, calculate the permutations starting with letters alphabetically before 'N'. Start with 'A'. For "AGAIN",- Starting with 'A', possible words are "AA, AG, AI".- For each "A" case (e.g., "AA, AG, AI") calculate permutations of the remaining letters.- For "AA", we have \( \frac{3!}{1!} = 6 \), ensures 6 words (AA...).- For "AG", \( \frac{3!}{1!} = 6 \), ensures 6 words (AG...).- For "AI", \( \frac{3!}{1!} = 6 \), ensures 6 words (AI...).We have used 18 words starting with 'A'.

Step 3: Move to the Next Letter in the Dictionary Order

Next, consider words starting with "G".- Similarly, 'GAI', 'GIA'. Start each case and calculate permutations of the remaining letters.- For "GA", \( \frac{3!}{1!} = 6 \), ensures 6 words (GA...).- For "GI", \( \frac{3!}{2!} = 3 \), ensures 3 words (GI...).This results in 9 more words starting with 'G', bringing our count to 27 words.

Step 4: Calculate Words Starting with Next Possible Character

Next, consider permutations starting with 'I'.- For each of these cases, calculate permutations left: - For "IA", \( \frac{3!}{1!} = 6 \), ensures 6 words (IA...). - For "IG", \( \frac{3!}{2!} = 3 \), ensures 3 words (IG...).Adding these 9 words, total is 36 (27 + 9).

Step 5: Determine Words Starting with 'N'

Since only 36 words are covered, need words starting with 'N'. Calculate next:- For words like "NA", follow similar permutation calculations: - For 'NAA', calculated permutations \( \frac{3!}{1!} = 6 \) and checked specifically arranged order.- Requires 13 conditions (find permutation count till 49th):- Specifically, 49th word or earlier ("NAAIG") confirmed satisfies based on repeating steps.

Key Concepts

Dictionary Order

Understanding dictionary order is like organizing words the way they appear in a dictionary. When we list permutations in dictionary order, each letter from a word is examined as if it's flipping pages in a dictionary.
For example, starting with 'A', we arrange all possible sequences beginning with that letter before moving on to the next.

• You start with the permutations that begin with the letters appearing earlier alphabetically.
• Continue with these permutations until all possible sequences are exhausted for that starting letter.

This ensures every permutation has a predetermined and consistent placement. So for the word 'AGAIN,' all the sequences would be explored in alphabetical order, just as you would look up words in a dictionary. This ordered arrangement helps in determining specific permutations like finding the 49th permutation.

Repeated Letters

The presence of repeated letters in a word, such as the two 'A's in 'AGAIN,' changes the number of unique permutations. When calculating permutations of a word with repeated letters, we need to adjust by dividing the total permutations by the factorial of the number of times each repeated letter appears.

• The formula for permutations of a word with repeated letters is \( \frac{n!}{p!} \).
• Here, \( n \) represents the total number of letters, and \( p \) is the factorial of the count of repeated letters.

For example, with 'AGAIN,' \( n = 5 \) (five letters total), and the count of repeated 'A's gives us \( p = 2! \). Applying the formula \( \frac{5!}{2!} \) gives 60 permutations. Without accounting for repetitions, one might overestimate the number of distinct arrangements.

Factorial Calculations

Factorials are mathematical expressions that represent the product of all positive integers up to a certain number. They are crucial in calculating permutations and combinations, especially with repeated letters.
Let's break it down:

• The factorial of a number \( n \), written as \( n! \), is the product of all positive integers from 1 to \( n \).
• For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

In permutation calculations, like determining total permutations of 'AGAIN,' the factorial helps determine how many ways the letters can be arranged. When letters repeat, we adjust by dividing \( n! \) by the factorial of the repeating letters to avoid counting identical arrangements more than once.
Hence, understanding factorials helps solve permutation problems efficiently while navigating through dictionary order and repeated letters nuances.