In how many different ways can the letters of the word 'DETAIL' be arranged such that all the vowels occupy only the odd positions (1st, 3rd and 5th positions)?

Introduction / Context:
This problem is another arrangement question with positional restrictions, similar to other vowel-consonant placement problems. The word DETAIL has six letters, including three vowels and three consonants. The condition is that the vowels must occupy only the odd positions in the arrangement. Since there are exactly three odd positions in a six-letter word and exactly three vowels, each odd position must be occupied by a vowel, and each even position by a consonant.

Given Data / Assumptions:
- Word: DETAIL.

- Letters: D, E, T, A, I, L.

- Vowels: E, A, I (3 vowels).

- Consonants: D, T, L (3 consonants).

- Positions: 6, labelled from 1 to 6.

- Odd positions: 1, 3, 5 (three positions).

- Even positions: 2, 4, 6 (three positions).

- Condition: vowels must occupy only odd positions, so positions 1, 3, 5 are for vowels, and positions 2, 4, 6 are consequently for consonants.

Concept / Approach:
Because the number of vowels equals the number of odd positions, and the condition forbids vowels in even positions, the placement of vowels and consonants is structurally forced: all vowels must be placed among the three odd positions, and all consonants must be placed among the three even positions. Within these slots, the vowels and consonants can be permuted independently. We count the permutations of vowels in their slots and multiply by the permutations of consonants in theirs.

Step-by-Step Solution:
Step 1: Identify the vowels: E, A, I and the odd positions: 1, 3, 5.Step 2: Since there are 3 vowels and 3 odd positions, each odd position will contain exactly one vowel.Step 3: The number of ways to arrange the 3 distinct vowels in positions 1, 3 and 5 is 3!.Step 4: Compute 3! = 3 * 2 * 1 = 6.Step 5: For the even positions (2, 4, 6), we must place the 3 consonants D, T, L.Step 6: The number of ways to arrange 3 distinct consonants in these 3 even positions is also 3! = 6.Step 7: Total valid arrangements = (ways to arrange vowels) * (ways to arrange consonants) = 3! * 3! = 6 * 6.Step 8: Compute 6 * 6 = 36.

Verification / Alternative check:
We can check that there is no alternative pattern: if any vowel were placed in an even position, the condition would be violated. Likewise, leaving any odd position for a consonant would force a vowel into an even position, again violating the rule. Hence there is a one-to-one correspondence between permutations of vowels and permutations of consonants, as captured in 3! * 3!. Recomputing 3! and the final product confirms the total of 36 arrangements.

Why Other Options Are Wrong:
- 25 and 42 are not products of two 3! terms and do not arise naturally from correct counting; they result from misusing combinations or missing a factor of permutations.

- 120 is the full 5! or a related count that ignores the positional restriction, effectively counting all or most permutations without separating vowels and consonants.

Common Pitfalls:
Some students misinterpret "vowels occupy only odd positions" as merely prohibiting vowels from even positions but allowing odd positions to be optionally occupied by consonants, which is impossible here because the counts of vowels and odd positions match exactly. Another mistake is to treat the task as a 6! problem and then attempt to divide by something, which does not reflect the structural constraints. Correctly recognising that there are two independent permutations—one for vowels and one for consonants—makes the problem straightforward.

Final Answer:
The letters of the word DETAIL can be arranged in 36 different ways so that all vowels occupy only the odd positions.