Problem 72 - 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) NAAGI (B) \(N A G A I\) (C) NAAIG (D) \(N A I A G\)

Step-by-Step Solution

Answer

The 49th word is 'NAAGI' (Option A).

Step 1: Understand the Problem

We need to find the 49th word in the dictionary arrangement of all permutations of the letters in the word 'AGAIN'. The key here is to arrange the letters in ascending order and count permutations until we reach the 49th permutation.

Step 2: List and Count Initial Arrangements

Sort the letters alphabetically: A, A, G, I, N. The total permutations are given by 5!/(2!) due to the two A's: \( \frac{5!}{2!} = 60 \) permutations.

Step 3: Start with 'A' as the First Letter

If 'A' is the first letter, the remaining letters A, G, I, N can form \( \frac{4!}{1!} = 24 \) permutations. Since 24 permutations with first letter 'A' are not enough to reach the 49th word, we move to the next possible starting letter.

Step 4: Consider 'G' as the First Letter

If 'G' is the first letter, the remaining letters A, A, I, N can form \( \frac{4!}{2!} = 12 \) permutations. Adding these to the permutations when starting with 'A', we have a total of 36 permutations so far (24 with 'A' and 12 with 'G'). This is still not enough to reach the 49th word.

Step 5: Move to 'I' as the First Letter

If 'I' is the first letter, the remaining letters A, A, G, N can form \( \frac{4!}{2!} = 12 \) permutations. Adding these 12 to the previous 36, we have reached 48 permutations (24 with 'A', 12 with 'G', and 12 with 'I') total.

Step 6: Find the First Word Starting with 'N'

Since 48 permutations account for starting with 'I', the 49th permutation starts with 'N'. The remaining letters A, A, G, I need to form a sequence. Arrange them: the smallest arrangement is A, A, G, I, thus the 49th permutation is 'NAAGI'.

Key Concepts

Dictionary Order

Understanding dictionary order is essential when arranging words composed of certain letters in "word order," just like how words are sorted in a dictionary.
In this exercise, the letters of the word "AGAIN" need to be organized as if they are words listed alphabetically in a dictionary.
Here's a simple guide to help you:

• First, arrange the letters in alphabetical order. From the word "AGAIN", we get A, A, G, I, N.
• Next, list all possible permutations or arrangements of these letters, starting with the smallest one as the first entry, moving on in a similar manner to the next least, and so on.
• This approach mirrors looking up a word in a dictionary. We always start at the top (e.g., A) and go down through other letters following the alphabet (e.g., G, I, N).

This systematic approach allows us to identify a specific position within this list.

Factorial Calculation

Factorial calculations are vital in determining how many unique arrangements can be made from a group of items, specifically letters in a word.
Let's delve into the meaning of factorial:

• A factorial of a number, say n, noted as n!, is the product of all positive integers up to n.
• For example, 3! (3 factorial) equals 3 × 2 × 1 = 6.

In contexts where certain letters repeat, we adjust our computation by dividing the result by the factorial of repeated letters' counts.
For "AGAIN", which has five letters, with A repeating twice, we compute permutations using

• \[ \frac{5!}{2!} \]
• 5! accounts for all arrangements of 5 letters, and 2! accounts for repetitions of "A".

Solving this gives us 60 possible arrangements.

Counting Permutations

Counting permutations helps us find specific sequences, like the 49th word in this context. Here's how it's done:
Begin by considering different letters as starting points, logically subdividing possible arrangements:

• First, count permutations starting with each possible letter in alphabetical order, beginning from the smallest to the largest.
• Prime examples in "AGAIN" include starting with A, then G, then I, effectively narrowing by each grouping of permutations.

For example, starting with "A" yields a smaller word grouping than "I" or "N". When you exhaust arrangements starting from smaller letters, move to the next in alphabetical order.
Repeated steps lead us near the desired sequence.
Once permutations reach 48 with the starting batches (like "A", "G", and "I"), "N" turns relevant. This reveals the actual 49th permutation quickly.

• Start with "N", then move down to a smaller subset: A, A, G, I.
• The whole process reveals the 49th permutation aptly as "NAAGI".