Question

The number of ways in which the letters of the word "BUSINESS" can be arranged so that the vowels always come together is

• A. 5!
• B. 6!
• C. 7!
• D. 8!

Hint:

When asked to arrange letters with some specific condition, treat the restricted group as a single block and then calculate the total arrangements.

Answer 1

The Correct Option is B

Step 1: The word BUSINESS has 8 letters: B, U, S, I, N, E, S, S, where the vowels are U, I, E and the letter S repeats three times among the consonants B, S, N, S, S.
Step 2: Glue the 3 vowels into one block, so there are 6 units to arrange: the vowel-block plus B, N, and three S's, giving \( \frac{6!}{3!} = 120 \) ways since S repeats three times.
Step 3: Multiply by the number of ways to arrange the 3 distinct vowels inside their own block, \( 3! = 6 \).
\[ 120 \times 6 = 720 = \boxed{6!} \]

Answer 2

Another way to count this is to build the arrangement by first placing the repeated consonant, then placing the remaining distinct units, rather than dividing by \( 3! \) at the end.

1. Treating the vowel block as one unit: Gluing U, I, E together into a single vowel-block leaves 6 units total to place in a row: the vowel-block, and the 5 consonants B, S, N, S, S.
2. Choosing positions for the repeated letter S: Among these 6 positions, the 3 identical S's must occupy 3 of the 6 slots. The number of ways to choose which 3 of the 6 positions hold an S is \( \binom{6}{3} = 20 \).
3. Placing the remaining distinct units: The 3 remaining slots must be filled by the 3 remaining distinct units, B, N, and the vowel-block, in some order, which can be done in \( 3! = 6 \) ways.
4. Arranging the vowels inside their block: Independently, the 3 distinct vowels inside the block can be ordered among themselves in \( 3! = 6 \) ways.

Multiplying all of these independent choices together: \( \binom{6}{3} \times 3! \times 3! = 20 \times 6 \times 6 = 720 \), which is exactly \( 6! \), matching the block-division method.

Therefore, the correct answer is 6!.