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:
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:
Show 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).
Updated On: Mar 25, 2026
- \( 34000 \)
- \( 37000 \)
- \( 36000 \)
- \( 35000 \)
Show Solution
Verified By Collegedunia
The Correct Option is C
Approach Solution - 1
To solve the problem of finding the number of words that can be formed using all the letters of the word "DAUGHTER" such that all the vowels never come together, we need to follow these steps:
- Identify the letters in the word "DAUGHTER". They are: D, A, U, G, H, T, E, R. There are a total of 8 letters.
- Count the vowels in "DAUGHTER". The vowels are A, U, E. There are 3 vowels.
- Calculate the total number of arrangements of the 8 letters:
\(8! = 40320\)
- Calculate the number of arrangements where the vowels all come together. Here, consider the 3 vowels (A, U, E) as one single entity or "super letter". This gives us new entities: (AUE), D, G, H, T, R.
- Count the number of these entities, which are 6 in number (5 consonants + 1 super letter of vowels).
- Calculate the arrangements of these 6 entities:
\(6! = 720\)
- Within the super letter (AUE), the vowels can be arranged among themselves. The number of such arrangements is:
\(3! = 6\)
- Thus, the total number of arrangements where all vowels come together is:
\(6! \times 3! = 720 \times 6 = 4320\)
- The required number of arrangements where vowels do not all come together is the total arrangements minus those where vowels are together:
\(8! - 6! \times 3! = 40320 - 4320 = 36000\)
- Thus, the number of words that can be formed such that all vowels never come together is 36,000.
Therefore, the correct answer is 36000.
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -2
Step 1: Total number of words.
The total number of words that can be formed using all the letters of the word "DAUGHTER" is:
\[
\text{Total words} = 8! = 40320.
\]
Step 2: Words with all vowels together.
The vowels in "DAUGHTER" are A, U, and E. If we treat these vowels as a single entity, the number of arrangements of the letters is:
\[
\text{Words with vowels together} = 6! \times 3! = 720 \times 6 = 4320.
\]
Step 3: Words with vowels not together.
To find the number of words where the vowels are not together, subtract the number of words with vowels together from the total number of words:
\[
\text{Words with vowels not together} = 8! - 6! \times 3! = 40320 - 4320 = 36000.
\]
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Permutation and Combination Questions
Let R = {(1, 2), (2, 3), (3, 3)}} be a relation defined on the set \( \{1, 2, 3, 4\} \). Then the minimum number of elements needed to be added in \( R \) so that \( R \) becomes an equivalence relation, is:
- The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together is:
- In a group of 3 girls and 4 boys, there are two boys \( B_1 \) and \( B_2 \). The number of ways in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but \( B_1 \) and \( B_2 \) are not adjacent to each other, is:
- The number of seven-digit numbers that can be formed by using the digits \(1,2,3,5,7\) such that each digit is used at least once, is:
- The number of ways of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions