In how many different ways can the letters of the word "ABACUS" be rearranged so that all the vowels appear together as a single block?

Introduction / Context:
This problem is another example of arranging letters with some of them treated as a group. The word "ABACUS" contains repeating letters and several vowels. We must count permutations where the vowels are together, which requires grouping them into a block and then handling repetition both within that block and in the overall arrangement.

Given Data / Assumptions:

• Word: ABACUS.
• Total letters = 6.
• Vowels: A, A, U (3 vowels with A repeated twice).
• Consonants: B, C, S (3 distinct consonants).
• All vowels must appear together as one contiguous block in every valid arrangement.
• All letters of the word are used in each arrangement.

Concept / Approach:
We use a block method with two stages. First, treat the group of vowels A, A and U as a single super letter or block. This block then joins the consonants B, C and S to form fewer effective objects. We count permutations of these objects. Second, we count the internal permutations of the vowels within the block, again compensating for the repeated A. The final answer is the product of these two counts.

Step-by-Step Solution:
Step 1: Treat the vowels A, A, U as a single block V. Now the objects are V, B, C and S. That means we have 4 distinct objects. Number of ways to arrange these 4 objects = 4! = 24. Step 2: Arrange the vowels inside the block V. The letters inside the block are A, A and U. Number of internal arrangements = 3! / 2! (since A repeats twice). Compute 3! = 6; divide by 2 to get 3. Total valid arrangements = (ways to arrange V, B, C, S) * (ways to arrange A, A, U). Total = 24 * 3 = 72.

Verification / Alternative check:
Check that the result is less than the total number of permutations of ABACUS without restriction. Without any grouping, total distinct permutations would be 6! / 2! because only A repeats. That equals 720 / 2 = 360. Our answer 72 is exactly one fifth of 360, which is reasonable since we force three letters to stick together. The method separates the external arrangement and internal arrangement correctly and respects the repeated letters, so we can be confident in 72 as the correct value.

Why Other Options Are Wrong:

• 12: This equals 4! / 2 and incorrectly treats the block or the repetition.
• 3: This is only the number of internal vowel arrangements and ignores the outer positions.
• 36: This is half of the correct value and usually comes from missing a factor of 2 somewhere in the counting.

Common Pitfalls:
Common errors include forgetting that the two A letters are identical, so dividing by 2! is necessary, or failing to treat the vowel group as a single block before applying the factorial count. Some learners also accidentally treat the block plus consonants as having fewer or more than 4 objects. Keeping the two stage structure clear and double checking the list of objects in each stage is the best way to avoid mistakes.

Final Answer:
The number of different arrangements of "ABACUS" in which all vowels appear together is 72.