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

Another way to check this is to compute the actual number instead of leaving it as a formula, then see which option gives that same number.

1. $6!/2!$: this is the total rearrangements of ABACUS with no restriction on the vowels at all, so it overcounts and equals 360, not what we need.
2. $3! \times 3!$: this equals 36, which is too small since it forgets to divide out the repeated A inside the vowel block.
3. $(3! \times 3!)/2!$: this equals 18, which under-counts because it treats only 3 outer units instead of 4.
4. $(4! \times 3!)/2!$: this equals 72, and this is the value we compute directly below.

Glue the vowels A, A, U into one block, giving 4 items to place: the block, B, C, S. That is $4! = 24$ orders. Inside the block, A, A, U can sit in $3! = 6$ orders, but swapping the two identical A's repeats a word, so divide by $2! = 2$, giving 3 distinct inner orders. Total distinct words = $24 \times 3 = 72$.

Let's summarize:

• Outer arrangement of 4 units: 24 ways.
• Inner arrangement of AAU: 3 distinct ways.
• Total: $24 \times 3 = 72$, matching $(4! \times 3!)/2!$.

This confirms option D by direct computation rather than by formula matching alone.