Question

In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

• A. \( \dfrac{6!}{2!} \)
• B. \( 3! \times 3! \)
• C. \( \dfrac{3! \times 3!}{2!} \)
• D. \( \dfrac{4! \times 3!}{2!} \)

Hint:

Glue the vowels A, A, U into one block, arrange it with the consonants, then divide out the repeated A inside the block.

Answer 1

The Correct Option is D

Step 1: List the letters and spot repeats.
ABACUS has 6 letters: A, B, A, C, U, S. The letter A repeats twice, and the vowels are A, A, U.

Step 2: Treat the vowel block as one unit.
Since the vowels must stay together, glue A, A, U into a single block. Now we arrange this block along with the consonants B, C, S, giving 4 units total: [AAU], B, C, S.

Step 3: Arrange the 4 units.
These 4 units can be arranged in \(4!\) ways.

Step 4: Arrange letters inside the vowel block.
Inside the block, A, A, U can be arranged in \( \dfrac{3!}{2!} \) ways, dividing by \(2!\) because the two A's are identical and swapping them gives the same word.

Step 5: Combine and match to an option.
Total arrangements = \( 4! \times \dfrac{3!}{2!} \). Writing it as one fraction gives \( \dfrac{4! \times 3!}{2!} \), which is option D. Options A, B, and C either ignore the repeated A altogether or arrange the wrong group of letters.

Final Answer:
The vowels-together arrangements of ABACUS total \( \dfrac{4! \times 3!}{2!} \). \[ \boxed{\dfrac{4! \times 3!}{2!}} \]