Key Concepts and Formulas

• Permutation: The number of ways to arrange n distinct objects is n! (n factorial).
• Complementary Counting Principle: The number of ways an event does not occur is the total number of outcomes minus the number of ways the event does occur.
• Treating a Group as One Unit: When certain items must be together, treat them as a single unit for arrangement purposes, and then multiply by the number of ways to arrange the items within that unit.

Step-by-Step Solution

Step 1: Analyze the word "DAUGHTER" and identify vowels and consonants.

• Why: We need to identify the letters in the word and classify them into vowels and consonants to apply the problem's constraints.
• Analysis:
  • The word "DAUGHTER" has 8 distinct letters: D, A, U, G, H, T, E, R.
  • Vowels: A, U, E (3 vowels)
  • Consonants: D, G, H, T, R (5 consonants)

Step 2: Calculate the total number of words that can be formed without any restrictions.

• Why: This is the total number of possible arrangements if we ignore the condition about vowels. It's the first part of using complementary counting.
• Calculation: Since all 8 letters are distinct, the total number of arrangements is simply 8!. $\text{Total number of words} = 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

Step 3: Calculate the number of words where all vowels are always together.

• Why: We need to determine the number of arrangements where the vowels are grouped together, as we'll subtract this from the total number of arrangements.
• Calculation:
  1. Treat the 3 vowels (A, U, E) as a single unit.
  2. Arrange the vowels within the unit: The 3 vowels can be arranged in 3! ways. $3! = 3 \times 2 \times 1 = 6$
  3. Arrange the units: Now we have 5 consonants (D, G, H, T, R) and 1 vowel unit (AUE), giving us a total of 6 units to arrange. These 6 units can be arranged in 6! ways. $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
  4. Multiply to get the total number of words with the vowels together: $\text{Words with vowels together} = 6! \times 3! = 720 \times 6 = 4320$

Step 4: Calculate the number of words where all vowels never come together.

• Why: Apply the complementary counting principle.
• Calculation: Subtract the number of words where vowels are together from the total number of words. $\text{Words with vowels never together} = \text{Total words} - \text{Words with vowels together}$ $= 8! - (6! \times 3!) = 40320 - 4320 = 36000$
  • Alternative (Efficient Factorization): $8! - 6! \times 3! = (8 \times 7 \times 6!) - (6! \times 3!) = 6! \times (8 \times 7 - 3!) = 6! \times (56 - 6) = 720 \times 50 = 36000$

Common Mistakes & Tips

• Factorials: Double-check your factorial calculations, especially when dealing with larger numbers.
• Internal Arrangements: Don't forget to account for the arrangements within the group that is treated as a single unit (e.g., the vowels).
• Distinct Letters: Always verify whether the letters in the given word are distinct or if repetitions exist. This will impact the formula used for permutations.

Summary

Using the complementary counting principle, we calculated the total number of possible words that can be formed from the letters of "DAUGHTER" (8!) and then subtracted the number of words where all the vowels are always together (6! * 3!). This gave us the number of words where the vowels never come together.

The final answer is $\boxed{36000}$, which corresponds to option (D).