Key Concepts and Formulas

• Permutations: The number of ways to arrange $n$ distinct objects is $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Principle of Complementary Counting: The number of ways to satisfy a condition is equal to the total number of ways minus the number of ways the condition is not satisfied.
• Block Method: When a group of objects must remain together, treat them as a single block.

Step 1: Analyze the Letters in the Word 'VOWELS'

What we are doing: Identify the vowels and consonants in the word 'VOWELS'. Why we do this: This is necessary to apply the problem's constraint (consonants not together) and use the block method effectively.

The word 'VOWELS' has 6 distinct letters: V, O, W, E, L, S. Categorizing them into vowels and consonants:

• Vowels: O, E (2 vowels)
• Consonants: V, W, L, S (4 consonants)

All letters are distinct, so we can use the permutation formula for distinct objects.

Step 2: Calculate the Total Number of Arrangements (Without Restrictions)

What we are calculating: The total number of unique six-letter words that can be formed using all the letters of 'VOWELS', without any conditions. Why we do this: This gives us our baseline – the universe of all possible arrangements from which we will subtract the unwanted cases.

Since there are 6 distinct letters, the total number of ways to arrange them is given by 6 factorial:

$\text{Total arrangements} = 6!$ $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

So, there are 720 distinct ways to arrange the letters of 'VOWELS'.

Step 3: Calculate the Number of Arrangements Where All Consonants Are Together

What we are calculating: The number of arrangements where the four consonants (V, W, L, S) always appear next to each other, forming a single group. Why we do this: This is the "unwanted" scenario according to the problem statement. By calculating this, we can subtract it from the total to find our desired answer.

To ensure all consonants stay together, we use the "block method":

1. Treat the consonants as a single unit (a 'block'): Imagine the four consonants (V, W, L, S) are glued together to form one super-letter: (VWLS).
2. Identify the units to be arranged: Now we have effectively 3 units to arrange:
   • The consonant block: (VWLS)
   • The first vowel: O
   • The second vowel: E
3. Arrange these units: These 3 units can be arranged among themselves in $3!$ ways. $3! = 3 \times 2 \times 1 = 6$
4. Arrange the letters within the consonant block: The four consonants (V, W, L, S) inside their block can also be arranged among themselves. Since there are 4 distinct consonants, they can be arranged in $4!$ ways. $4! = 4 \times 3 \times 2 \times 1 = 24$
5. Combine the arrangements: To get the total number of arrangements where all consonants are together, we multiply the number of ways to arrange the units by the number of ways to arrange the letters within the consonant block (by the multiplication principle).

$\text{Arrangements with consonants together} = (\text{Arrangements of units}) \times (\text{Arrangements within consonant block})$ $\text{Arrangements with consonants together} = 3! \times 4! = 6 \times 24 = 144$

So, there are 144 arrangements where all the consonants are together.

Step 4: Calculate the Number of Arrangements Where All Consonants Are Not Together

What we are calculating: The number of arrangements where the consonants are never together. Why we do this: Applying the principle of complementary counting.

$\text{Arrangements (consonants not together)} = \text{Total arrangements} - \text{Arrangements (consonants together)}$ $\text{Arrangements (consonants not together)} = 720 - 144 = 576$

Final Answer

The number of six-letter words formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is $\boxed{576}$. This does not match any of the options. The provided correct answer is incorrect.