The word 'Director' has 8 letters with R repeated twice (letters: D, I, R, E, C, T, O, R). In how many distinct arrangements of all letters are the three vowels (I, E, O) not all together in one single block?

Introduction / Context:
This is a permutations-with-repetition problem. The word “Director” contains 8 letters with a repetition of R (twice). We need the number of arrangements where the three vowels (I, E, O) are not all together as one contiguous block. Note this condition does not forbid two vowels from being adjacent; it only excludes the case where all three vowels appear consecutively as one block.

Given Data / Assumptions:

• Letters: D, I, R, E, C, T, O, R.
• Vowels: I, E, O (distinct).
• Consonants: D, C, T, R, R (R repeats).
• All 8 letters are used in each arrangement.

Concept / Approach:
Total arrangements minus arrangements with all three vowels together (as one block). This classical “complement” strategy is simpler than directly counting all acceptable patterns.

Step-by-Step Solution:
Total arrangements (accounting for repeated R): 8! / 2! = 20160.Count arrangements with vowels together as one block: Treat (I,E,O) as a single super-letter. Then we have 6 entities: [Vowel-Block] + D + C + T + R + R.Ways to arrange these 6 with two R’s: 6! / 2! = 360.Internal arrangements of vowels within the block: 3! = 6.Arrangements with all three vowels together: 360 * 6 = 2160.Required arrangements = 20160 − 2160 = 18000.

Verification / Alternative check:
Total 20160 is standard for 8 letters with a double letter. The “vowels-together” count uses block method with repeated consonant R handled by division by 2!.

Why Other Options Are Wrong:
20160 counts all permutations; 16000 and 1800/1600 are not supported by the block-minus-complement calculation.

Common Pitfalls:
Misreading “never together” to mean “no two vowels adjacent.” Here the stem explicitly says “the three vowels are never together,” i.e., exclude only the single-block case.

Final Answer:
18000