INDIA Word Permutation Analysis

We need to find the number of ways to arrange the letters of the word INDIA so that the vowels appear only in the odd positions. Let's first understand the components of the word INDIA:

• The word INDIA has 5 letters. We will assume these 5 letters are distinct for the purpose of this permutation calculation (e.g., I, N, D, R, A).
• Vowels in the word are: I, A. There are 2 vowels.
• Consonants in the word are: N, D, R. There are 3 consonants.

Identifying Available Positions

A 5-letter arrangement requires 5 positions. We need to identify which ones are odd and which are even:

• The positions are numbered 1, 2, 3, 4, 5.
• The odd positions are 1, 3, and 5. There are 3 odd positions available.
• The even positions are 2 and 4. There are 2 even positions available.

Vowel Placement Calculation

The main condition is that the vowels must be placed in the odd positions. We have 2 vowels (I, A) and 3 available odd positions (1, 3, 5).

• We need to find the number of ways to arrange these 2 vowels within the 3 odd positions. Since the order of vowels matters (e.g., IA is different from AI), this is a permutation problem.
• The formula for permutations is \( P(n, k) = \frac{n!}{(n-k)!} \), where \( n \) is the total number of available positions, and \( k \) is the number of items to arrange.
• In this scenario, \( n = 3 \) (number of odd positions) and \( k = 2 \) (number of vowels).
• Let's calculate the number of ways: \( P(3, 2) = \frac{3!}{(3-2)!} \)
• \( P(3, 2) = \frac{3!}{1!} \)
• Since \( 3! = 3 \times 2 \times 1 = 6 \) and \( 1! = 1 \), the calculation becomes: \( P(3, 2) = \frac{6}{1} = 6 \)

Therefore, there are 6 distinct ways to arrange the vowels I and A within the available odd positions (1st, 3rd, 5th).

Consonant Placement Context

To complete the arrangement, the 3 consonants (N, D, R) need to be placed in the remaining \( 5 - 2 = 3 \) positions. These remaining positions consist of the two even slots (2nd, 4th) and the one odd slot that was not occupied by a vowel. The number of ways to arrange these 3 consonants in the 3 remaining spots is \( P(3, 3) = 3! = 6 \). The total number of permutations for the word INDIA with vowels in odd positions would be the product of the ways to arrange vowels and the ways to arrange consonants (\( 6 \times 6 = 36 \)). However, the calculation specifically for placing vowels in odd positions yields 6.