Question

60 is t he t ot al number of words wit h or wit hout definit ion which can be made usingall let t ers of t he word again. figure-out t he 49t h word if t hese words arement ioned as in wordbook?

a. NAGIA
b. NAIGA
c. NAAIG
d. NAAGI

Answer 1

The 49th word is NAAGI and the correct option is D.

To determine this, we can break down the problem into smaller steps:

1. Determine the total number of permutations of the word "AGAIN".

There are five letters in the word "AGAIN": A, G, A, I, and N. Since the word has two A's, we have to consider permutations with and without repetition.

Permutations with repetition:

For each letter, there are five choices (since we can use any of the five letters). So, the total number of permutations with repetition is 5^5 = 3125.

Permutations without repetition:

Since there are two A's, we can treat them as distinct letters initially. So, we have 5 distinct letters: A, G, A, I, and N. The number of permutations without repetition is 5! = 5×4×3×2×1 = 120.

However, we have overcounted the permutations since the two A's are not distinct. To correct for this, we divide by 2! (the number of permutations of the two A's): 120/2! = 60.

Therefore, the total number of permutations is 60.

2. Identify the 49th word alphabetically.

The 49th word will start with the letter 'N' since there are only 48 words that start with 'A', 'G', or 'I'.

The remaining letters can be arranged in 4! = 24 ways.

So, the 49th word is NAAGI, followed by NAAIG and the correct option is D.