The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all t

Q: Wait, why do we subtract the 'together' cases from the total? Isn't that just for 'at least one' problems?

That is a great observation! The Complementary Principle is a universal tool. The condition 'vowels never come together' is the exact opposite of 'all vowels are together'. By finding the total universe of arrangements and removing the specific 'together' scenario, we are left with exactly the set of arrangements where they are separated. It is much faster than trying to place consonants and then slotting vowels into the gaps!

Q: Why do we multiply by $3!$ in the block method? I thought the block was just one unit?

Think of the block as a container. While the container itself moves as one unit among the consonants, the contents inside (A, U, E) can shuffle around. Since there are 3 distinct vowels, they can be arranged in $3!$ (which is $3 \times 2 \times 1 = 6$) different ways inside that container. If you forget this, you are missing 5 out of every 6 possible 'together' arrangements!

Q: Can I use the Gap Method instead? It feels more intuitive.

Absolutely! The Gap Method is a fantastic alternative. You would arrange the 5 consonants first ($5!$ ways), which creates 6 gaps (including the ends). Then, you choose 3 gaps out of 6 to place your vowels ($^{6}P_{3}$ ways). The result will be $5! \times ^{6}P_{3} = 120 \times 120 = 14400$. Wait—let me re-check that. Actually, the Complementary Method is often less prone to calculation errors for this specific problem. Both methods are valid, but always double-check your arithmetic!

Question

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

Options:

(A) 36000

(B) 37000

(C) 34000

(D) 35000

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The word **DAUGHTER** consists of 8 distinct characters: $\{D, A, U, G, H, T, E, R\}$. Among these, the vowels are $\{A, U, E\}$ and the consonants are $\{D, G, H, T, R\}$.

Our objective is to determine the number of arrangements where the vowels never appear together.

### The Total Universe

First, we calculate the total number of ways to arrange all 8 distinct letters without any restrictions. This is given by the permutation of $n$ distinct objects, which is $n!$.

For 8 letters, the total number of arrangements is:

$$8! = 8 	imes 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 40320$$

This value represents our total sample space.

### The Complementary Strategy

To find the number of arrangements where the vowels are *never* together, we employ the **Complementary Method**. We subtract the number of arrangements where the vowels *are* together from the total number of arrangements.

$$	ext{Required Arrangements} = (	ext{Total Arrangements}) - (	ext{Arrangements where vowels are together})$$

### The Block Method

To calculate the cases where the vowels are together, we treat the set $\{A, U, E\}$ as a single "block" or unit. We now have 5 consonants and 1 vowel block, totaling 6 units to arrange.

The number of ways to arrange these 6 units is:

$$6! = 720$$

### The Internal Dance

Within the vowel block, the letters $\{A, U, E\}$ can be arranged among themselves in $3!$ ways.

$$3! = 3 	imes 2 	imes 1 = 6$$

By the Fundamental Principle of Counting, the total number of arrangements where the vowels are together is:

$$720 	imes 6 = 4320$$

### Final Calculation

We now subtract the forbidden cases from the total universe to arrive at the final result.

$$40320 - 4320 = 36000$$

The number of ways to arrange the letters of the word **DAUGHTER** such that the vowels never sit together is **36000**.

Animated Solution for Mathematics - Permutations and Combinations: The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is

Select Answer:

Visualized Solution

Analyze the word DAUGHTER

Total Possible Arrangements

The Complementary Strategy

The Block Method

Arranging the $6$ Units

Internal Arrangement of Vowels

Total Vowels Together Cases

Final Subtraction: The Result

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Total Universe

The Complementary Strategy

The Block Method

The Internal Dance

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Permutations and Combinations: The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is

Select Answer:

Visualized Solution

Analyze the word DAUGHTER

Total Possible Arrangements

The Complementary Strategy

The Block Method

Arranging the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6 Units

Internal Arrangement of Vowels

Total Vowels Together Cases

Final Subtraction: The Result

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Total Universe

The Complementary Strategy

The Block Method

The Internal Dance

Final Calculation

Similar Questions

The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER" in which vowels never come together is

The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is

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 .

The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ____.

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is

How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently is

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

Arranging the $6$ Units