The number of six letter words (with or without meaning), formed using all the letters of the word '

Q: Why can't I just use the gap method to place the consonants?

The gap method is excellent when you want to ensure NO two consonants are adjacent. However, the question asks for the case where all consonants NEVER come together as a single block. This is a subtle but crucial difference! The gap method would be overkill and incorrect here because it restricts the consonants more than the problem requires.

Q: Does 'all consonants never come together' mean no two consonants can be adjacent?

No, it does not! It simply means the specific scenario where all four consonants (V, W, L, S) form a single contiguous block is forbidden. Other arrangements, like 'V W O L S E', are perfectly valid because the consonants are not all together in one group.

Q: Why do we multiply by $4!$ at the end?

Great question! When we treat the four consonants as a single unit, we are arranging the 'block' relative to the vowels. But inside that block, the consonants V, W, L, and S can shuffle among themselves in $4!$ different ways. We must account for these internal permutations to get the total count of 'together' cases.

Question

The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is .

Sigma Percentile · Accepted Answer

### The Art of Counting: Mastering Permutations

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, logical structure beneath.

We are looking at the word 'VOWELS' and asking ourselves: how many ways can we arrange these letters such that all the consonants never come together? This is not just a math problem; it is a lesson in strategic thinking.

### Phase 1: The Anatomy of 'VOWELS'

First, let us observe our raw material. The word is 'VOWELS'. We have a total of $n=6$ letters.

If we look closely, we can partition these into two distinct groups: the vowels $\{O, E\}$ and the consonants $\{V, W, L, S\}$. We have $2$ vowels and $4$ consonants. This classification is the first step in any combinatorial journey.

### Phase 2: The Total Universe

Before we impose any restrictions, let us imagine the entire universe of possibilities. If we were to arrange these $6$ distinct letters without any conditions, the number of ways is simply $6!$.

Calculating this, we get:
$$6! = 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 720$$

This is our total sample space. Every valid arrangement we seek must exist within this set of $720$ possibilities.

### Phase 3: The Complementary Strategy

Now, we face the condition: 'all the consonants never come together'. If we tried to count this directly, we would have to account for complex casework.

Instead, we use the elegant **Complementary Method**. We calculate the total arrangements and subtract the cases where the condition is *violated*.

The violation here is the scenario where all four consonants *do* come together. Our strategy is:
$$	ext{Required} = 	ext{Total} - 	ext{Together}$$

### Phase 4: The String Method

To find the 'together' cases, we use the powerful **String Method** (or 'Block Method'). We imagine tying the four consonants $\{V, W, L, S\}$ together with an invisible string, treating them as one single, unbreakable unit.

Now, look at what remains: we have this one 'consonant block' and the two individual vowels $\{O, E\}$. We are now arranging $1 + 2 = 3$ units.

The number of ways to arrange these $3$ units is:
$$3! = 6$$

### Phase 5: The Internal Dance

We are not done yet! Within that 'consonant block', the letters $V, W, L,$ and $S$ are not static. They can shuffle among themselves.

The number of ways to arrange these $4$ distinct consonants internally is $4!$. Calculating this, we get:
$$4! = 4 	imes 3 	imes 2 	imes 1 = 24$$

Therefore, for every arrangement of our $3$ units, there are $24$ internal variations of the consonant block.

### Phase 6: The Final Calculation

To find the total number of 'together' cases, we multiply the arrangements of the units by the internal arrangements of the block:
$$3! 	imes 4! = 6 	imes 24 = 144$$

Finally, we return to our complementary strategy. We take our total universe of $720$ and subtract the $144$ cases where the consonants are together:
$$720 - 144 = 576$$

There you have it—the elegance of logic in action. You have successfully navigated the trap and arrived at the final answer: **576**.

Animated Solution for Mathematics - Permutations and Combinations: The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is .

Enter Numerical Value:

Visualized Solution

Analyze the Word 'VOWELS'

Calculate Total Arrangements

The Complementary Method

The String/Block Method

Arranging the Units

Internal Arrangement of Consonants

Total 'Together' Cases

Final Subtraction

The Sigma Insight: Linear Permutations

The Art of Counting

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Phase 6

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