In how many different ways can the letters of the word 'OLIVER' be arranged so that the vowels in the word always appear in alphabetical (dictionary) order from left to right?

Introduction / Context:
This problem focuses on permutations of letters with a restriction on the relative order of vowels. The word OLIVER contains six distinct letters, including three vowels. The condition is that whenever the vowels appear from left to right in any arrangement, they must follow dictionary order, meaning E must come before I, and I must come before O. Their positions in the word may vary, but their left-to-right order must always be E, then I, then O. This type of constraint is common in permutation questions to test understanding of relative ordering.

Given Data / Assumptions:
- Word: OLIVER.
- Total letters: 6 distinct letters (O, L, I, V, E, R).
- Vowels: O, I, E (3 vowels).
- Consonants: L, V, R (3 consonants).
- We must count permutations where, reading from left to right, the vowels appear in alphabetical order E, I, O, although other letters can appear in between.

Concept / Approach:
Without any restriction, all 6 letters can be arranged in 6! different ways. In those unrestricted permutations, the 3 vowels can appear in any of 3! possible relative orders with equal frequency. However, only 1 of these orders (E, I, O) satisfies the dictionary order condition. Therefore, the number of valid permutations is the total permutations divided by the number of possible vowel orders, effectively 6! / 3!. This is a powerful symmetry argument that avoids direct position-by-position counting.

Step-by-Step Solution:
Step 1: Compute the total number of permutations of 6 distinct letters with no restriction.Step 2: Total permutations = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720.Step 3: Among these permutations, the vowels E, I, O can appear in 3! = 6 different relative orders: EIO, EOI, IEO, IOE, OEI, OIE.Step 4: Each of these 6 vowel-order patterns appears in exactly the same number of permutations, because all letters are distinct and symmetric in the counting.Step 5: We want only the permutations where vowels appear in the single order E, I, O.Step 6: Therefore, valid permutations = total permutations / number of vowel orders = 6! / 3! = 720 / 6.Step 7: Compute 720 / 6 = 120.

Verification / Alternative check:
An alternative method is to choose positions for the three vowels and then place the consonants, explicitly ensuring that the vowel positions follow E, I, O. First choose any 3 positions out of 6 for the vowels, which can be done in 6C3 ways. For each such choice, the vowels must occupy these positions in the fixed order E, I, O, so there is only 1 way to place vowels. The remaining 3 positions are filled with 3 consonants in 3! ways. Thus, total arrangements = 6C3 * 3! = 20 * 6 = 120, matching the earlier argument.

Why Other Options Are Wrong:
- 144 and 136 are close but arise from incorrect combinatorial factors or incomplete use of factorials.
- 186 is not consistent with any natural factorisation like 6C3 * 3! or 6!/3! for this situation and therefore does not match the proper counting logic.

Common Pitfalls:
Students sometimes enforce that the vowels must be adjacent or occupy specific positions, which is not required here; only their relative order matters. Others wrongly count only one vowel ordering and forget to divide the total by 3!, or they attempt to overcomplicate the positional selection. Using symmetry and the idea of equally likely vowel orderings is the cleanest and most reliable approach.

Final Answer:
The letters of the word OLIVER can be arranged in 120 different ways so that the vowels appear in dictionary order from left to right.