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: Treating vowels as a block.
The word "BUSINESS" contains the vowels U, I, and E. To ensure that the vowels always come together, we treat them as a single unit or block. So, we have the following letters left: B, S, N, S, and the block of vowels (UIE).
Step 2: Arranging the blocks.
Now, we have 6 units: B, S, N, S, and the vowel block. The number of ways to arrange these 6 units is \( \frac{6!}{2!} \), because the letter S is repeated twice.
Step 3: Arranging the vowels.
The 3 vowels (UIE) can be arranged among themselves in \( 3! \) ways.
Step 4: Total number of arrangements.
The total number of ways to arrange the letters so that the vowels are together is: \[ \frac{6!}{2!} \times 3! = \frac{720}{2} \times 6 = 7! = 5040 \]
Step 5: Conclusion.
Therefore, the correct answer is (3) 7!.

Answer 2

This question asks for the number of arrangements of the letters of "BUSINESS" in which the three vowels stay together, so it's worth first getting the letter inventory exactly right: the word has 8 letters, B, U, S, I, N, E, S, S, meaning the vowels are U, I, E (three distinct vowels) and the consonants are B, S, N, S, S, where the letter S is repeated three times among the consonants.

1. 5!: This is far too small for this problem; it would only correspond to arranging 5 distinct units with no repetition at all, which does not match the actual letter counts involved.
2. 6!: Treat the three vowels U, I, E as a single glued block. Together with the 5 consonants B, S, N, S, S, this gives 6 units to arrange: the vowel-block, B, N, and the three S's. Since S repeats three times among these 6 units, the number of distinct ways to arrange them is \( \frac{6!}{3!} = \frac{720}{6} = 120 \). The 3 vowels inside their block can be arranged in \( 3! = 6 \) ways. Multiplying gives \( 120 \times 6 = 720 = 6! \), so this option is consistent with the correct count.
3. 7!: This value would arise from mistakenly treating S as repeating only twice instead of three times, understating the repetition and producing an inflated count.
4. 8!: This is the count for arranging all 8 letters with no restriction and without accounting for the repeated S at all, overcounting on two fronts.

Correctly identifying that S occurs three times among the consonants, and combining the block-arrangement count with the internal vowel-arrangement count, gives \( 6! \) as the true number of valid arrangements.

Therefore, the correct answer is 6!.