In how many different ways can the letters of the word 'DESIGN' be arranged so that no consonant appears in either the first or the last position (that is, the two end positions must both be vowels)?

Introduction / Context:
This problem involves arranging letters of a word with restrictions on the end positions. The word DESIGN consists of six distinct letters and a mix of vowels and consonants. The condition is that no consonant may appear at either end of the word. Therefore, both the first and last positions must be occupied by vowels. The remaining positions can be filled by consonants without any extra restrictions. This tests your ability to separate positional constraints from the general permutation.

Given Data / Assumptions:
- Word: DESIGN.

- Letters: D, E, S, I, G, N (6 distinct letters).

- Vowels: E, I (2 vowels).

- Consonants: D, S, G, N (4 consonants).

- Positions: 6, labelled from 1 to 6.

- Condition: positions 1 and 6 (the two ends) must not be consonants, so they must be vowels.

Concept / Approach:
Because there are only two vowels and exactly two end positions, both vowels must occupy the first and last positions in some order. Once we place E and I in positions 1 and 6, the remaining 4 positions in the middle can be filled freely by the 4 consonants D, S, G, N. Since all letters are distinct, this becomes a straightforward multiplication of the permutations for vowels at the ends and permutations for consonants in the middle positions.

Step-by-Step Solution:
Step 1: Identify which letters are vowels and which are consonants.Step 2: Vowels: E and I; consonants: D, S, G, N.Step 3: The two end positions (first and last) must be occupied by vowels.Step 4: There are 2 vowels and 2 end positions, so the vowels can be arranged in these positions in 2! ways (E at first and I at last, or I at first and E at last).Step 5: The remaining 4 positions in the middle must be filled with the 4 consonants D, S, G, N.Step 6: All 4 consonants are distinct, so they can be arranged in the 4 middle positions in 4! ways.Step 7: Compute 4! = 4 * 3 * 2 * 1 = 24.Step 8: Total valid arrangements = (ways to arrange vowels at the ends) * (ways to arrange consonants in the middle) = 2! * 4! = 2 * 24 = 48.

Verification / Alternative check:
We can confirm that no other pattern is possible because there are exactly as many vowels as there are end slots. If one of the end positions were filled by a consonant, we would have to place at least one vowel in the middle, violating the condition that no consonant appears at either end. Therefore, the only valid structure is vowel–middle consonants–vowel, which our counting fully captures. Recomputing 2 * 24 also confirms the arithmetic as 48.

Why Other Options Are Wrong:
- 240 and 72 are larger than 48 and likely come from ignoring one of the constraints, such as treating the end positions as free or not separating vowel and consonant roles correctly.

- 36 is smaller than 48 and might result from mistakenly dividing by 2 or 4, perhaps due to confusion over repeated letters (even though there are none here).

Common Pitfalls:
A common mistake is misidentifying vowels and consonants or assuming that vowels can also appear in the middle with some unused at the ends. But since only vowels can occupy the ends and there are precisely two vowels, all vowel positions are forced. Another error is forgetting that the order of consonants in the middle matters, leading to an undercount by using combinations instead of permutations.

Final Answer:
The letters of the word DESIGN can be arranged in 48 ways so that no consonant appears at either end.