In how many different ways can the letters of the word AUCTION be arranged in such a way that the vowels always come together?

### Concept & Permutations with Grouping To ensure specific items are always adjacent, group them into a single super-entity. The total arrangements equal the permutations of all entities multiplied by the internal permutations of the super-entity. ### Step-by-Step Solution * **Analyze the Word:** The word "AUCTION" contains 7 letters. * **Identify Vowels and Consonants:** * Vowels: A, U, I, O (4 letters) * Consonants: C, T, N (3 letters) * **Group the Vowels:** Treat the 4 vowels (A, U, I, O) as a single unit. * **Count the Units:** We have the consonants C, T, N and the vowel unit (AUIO), making $3 + 1 = 4$ units in total. * **Arrange the Units:** The 4 units can be arranged in $4!$ ways. * $4! = 24$ * **Arrange Items Within the Unit:** The 4 vowels can be arranged internally in $4!$ ways. * $4! = 24$ * **Calculate Total Arrangements:** Multiply the independent arrangements. * Total ways = $24 \times 24 = 576$ ### Exam Strategy & Shortcut Recognize that neither the vowels nor the consonants have repeating letters. Apply the direct logical formula: $(C+1)! \times V!$, where $C = 3$ and $V = 4$. So, $4! \times 4! = 576$. ### Common Pitfall Mistaking the vowel count. Missing one vowel while scanning the word (like missing the 'I' or 'O') changes the calculation entirely. Always cross-check your letter tally with the original word length. ### Final Answer Therefore, the correct answer is **576**.