Question

The number of words that can be formed using all the letters of the word "DAUGHTER" such that all the vowels never come together, is:

• A. \( 34000 \)
• B. \( 37000 \)
• C. \( 36000 \)
• D. \( 35000 \)

Hint:

When counting arrangements where certain items must not be together, first count the total arrangements, then subtract the unwanted cases (where the items are together).

Answer 1

The Correct Option is C

To determine the count of distinct arrangements of the letters in "DAUGHTER" where all vowels are not adjacent, follow these computational steps:

1. List the letters of "DAUGHTER": D, A, U, G, H, T, E, R. This yields a total of 8 unique letters.
2. Identify and count the vowels within "DAUGHTER": A, U, E. There are 3 vowels.
3. Compute the total permutations for all 8 letters:

\(8! = 40320\)

1. Calculate arrangements where the vowels (A, U, E) are treated as a single composite unit. The distinct units become: (AUE), D, G, H, T, R.
2. Determine the total number of these new entities, which is 6 (5 consonants + 1 vowel cluster).
3. Calculate the permutations of these 6 entities:

\(6! = 720\)

1. Within the vowel cluster (AUE), the vowels can be internally arranged. The number of such internal arrangements is:

\(3! = 6\)

1. Multiply the permutations of the entities by the internal permutations of the vowels to find arrangements where all vowels are together:

\(6! \times 3! = 720 \times 6 = 4320\)

1. Subtract the arrangements where vowels are together from the total arrangements to find those where vowels are not all together:

\(8! - 6! \times 3! = 40320 - 4320 = 36000\)

1. The final result indicates that 36,000 distinct words can be formed such that no two vowels are adjacent.

The definitive count is 36000.