Problem 35 - A Complete Resource Book in Mathematics for JEE Main 2017

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) \(\mathrm{NAAGI}\) (B) \(\mathrm{NAGAI}\) (C) NAAIG (D) \(\mathrm{NAIAG}\)

Step-by-Step Solution

Answer

The 49th word is NAAGI.

Step 1: Calculate Total Permutations

To find the total number of permutations of the word 'AGAIN', calculate using the formula for permutations of letters with repetition. The word has 5 letters with 'A' repeating twice: \[ \frac{5!}{2!} = \frac{120}{2} = 60 \] So, there are 60 permutations.

Step 2: Determine Permutation Groupings

Group these permutations alphabetically using 'A' at the start. Since there are 2 'A's, we determine permutations starting with each letter separately:- Starting with 'A': Letters left are 'A', 'G', 'I', 'N' = Total: \(\frac{4!}{1!} = 24\)- Starting with 'G': Letters are 'A', 'A', 'I', 'N' = Total: \(\frac{4!}{2!} = 12\)- Starting with 'I': Same as 'G', total is also 12.- Starting with 'N': Same group as 'G' and 'I', total is also 12.

Step 3: Finding the 49th Permutation

Starting from the 37th position, we look for the 49th permutation: - 'A' category covers positions 1-24. - 'G' category covers positions 25-36. - 'I' category covers positions 37-48. - Therefore, position 49 is in the 'N' category.

Step 4: Identify Permutations Starting with 'N'

Within the 'N' group, which starts at position 49: - Remaining letters are 'A', 'A', 'G', 'I'. - Arrange them lexicographically: NAAIG, NAAGI, NAGAI, NAGIA, NAIAG, NAIGA... Hence, the 49th word is NAAGI.

Key Concepts

Dictionary Order of Permutations

When we talk about arranging words or sets of letters in dictionary order, it means we are putting them in the same order you would find in a dictionary. This method orders words based on the alphabetical order of their letters.
For example:

• If we have the letters 'B', 'A', 'C', the dictionary order would be 'ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA'.

The dictionary order helps us determine where a specific permutation falls within a sequence when arranged alphabetically.
In the context of permutations, using dictionary order allows us to systematically list every possible arrangement and easily identify the position of each permutation within this ordered list.
For our exercise, understanding the dictionary order is crucial. Knowing the order makes it easier to know which group of permutations a particular word falls into.

Factorial Calculation

Factorial calculation is an essential concept when dealing with permutations. Notation for factorial is an exclamation mark.

• For example, the factorial of 5 is written as 5! and calculated as 5 x 4 x 3 x 2 x 1, which equals 120.

In combinatorics, factorial helps to find the total number of ways to arrange a set of items.
For example:

• With 5 unique items, we can arrange them in 5! or 120 different ways.

Factorial calculation becomes slightly adjusted when repeated items are involved, as with the word "AGAIN", where the letter 'A' repeats. The formula adjusts for these repetitions to avoid overcounting:

• For 'AGAIN', it's: \( \frac{5!}{2!} \), which equals 60.

This accounts for the two 'A's and ensures that we only get valid distinct permutations.

Handling Repeated Letters

Repeated letters in a word influence the permutation count because identical items do not create new unique sequences when swapped.
In permutation calculation:

• Repeated letters require dividing by the factorial of the count of those repeated letters.

Let's revisit "AGAIN", which contains two 'A's. Using the permutation formula for repetition is necessary:

• \( \frac{5!}{2!} \) equals 60.

Here, \(2!\) in the denominator corrects for the repetition of 'A'.
If there were more repetitions (say three of one letter in "AAAIG"), the adjustment in the formula would involve that value's factorial:

• For three identical letters, \( \frac{5!}{3!} \) would be needed to calculate valid permutations.

Managing repeated letters correctly is crucial to getting accurate counts of distinct permutations.