How many arrangements of the letters of the word 'BENGALI' can be made if all the vowels appear only in odd positions of the arrangement (and never in even positions)?

Introduction / Context:
This question combines permutations with position-based restrictions. The word BENGALI has seven distinct letters, with some vowels and some consonants. The restriction is that vowels must occupy only odd positions in the final arrangement, although odd positions may also contain consonants if some odd spots remain unused by vowels. This typical pattern appears frequently in placement and entrance exams to test understanding of counting under positional conditions.

Given Data / Assumptions:
- Word: BENGALI.
- Total letters: 7, all distinct (B, E, N, G, A, L, I).
- Vowels: E, A, I (3 vowels).
- Consonants: B, N, G, L (4 consonants).
- Positions available: 7, labeled 1 to 7.
- Odd positions: 1, 3, 5, 7 (4 positions).
- Even positions: 2, 4, 6 (3 positions).
- Vowels must occupy only odd positions; they cannot appear in even positions.

Concept / Approach:
The vowels must be placed in odd positions only. There are 4 odd positions and 3 vowels, so we must choose 3 of the 4 odd positions for the vowels, then arrange the vowels among those chosen positions. After that, we place the 4 consonants in the remaining 4 positions (3 even positions plus the 1 unused odd position). Each stage is counted using combinations or permutations as appropriate, and the total is the product of these counts.

Step-by-Step Solution:
Step 1: Choose which 3 of the 4 odd positions will be occupied by vowels.Step 2: The number of ways to choose 3 positions out of 4 is 4C3 = 4.Step 3: Arrange the 3 vowels (E, A, I) in the chosen 3 positions. Since they are distinct, there are 3! = 6 ways.Step 4: Now 4 positions remain (the 3 even positions plus the 1 unused odd position), and we must place the 4 consonants (B, N, G, L) there.Step 5: The 4 consonants can be arranged in these 4 positions in 4! = 24 ways.Step 6: Total valid arrangements = 4C3 * 3! * 4! = 4 * 6 * 24.Step 7: Compute: 4 * 6 = 24, and 24 * 24 = 576.

Verification / Alternative check:
You can also reason by first placing vowels and then consonants without explicitly splitting between odd and even counts. The number of ways to assign vowels to odd positions and then fill the rest with consonants remains 4C3 * 3! * 4!, which yields 576. Trying a small example with fewer letters and positions can help confirm that this logic scales correctly to the full problem, increasing confidence in the computation.

Why Other Options Are Wrong:
- 720 is 6! or a related value and typically arises from forgetting the restriction on positions and counting all permutations or misusing factorials.
- 567 is a near miss that might result from arithmetic slips rather than a correct combinatorial argument.
- 625 does not factor naturally into products of small factorials and combinations for this problem and does not reflect any standard counting step here.

Common Pitfalls:
One frequent mistake is to assume that all odd positions must be occupied by vowels, which would be impossible here because there are 4 odd positions but only 3 vowels. Another error is to place vowels in any positions and then try to subtract disallowed cases, which makes the counting complicated and error-prone. Using the direct constructive approach—choose allowed positions for vowels and then permute consonants in the remaining slots—keeps the logic clear and avoids overcounting.

Final Answer:
The letters of the word BENGALI can be arranged in 576 different ways so that vowels appear only in odd positions.