JEE Mathematics 2025 Question with Solution

Given: The word is "DAUGHTER" and all its letters are distinct. Find: The number of words formed using all letters such that all the vowels never come together.

Total number of letters is $8$. The vowels are $A, U, E$, so the number of vowels is $3$.

First count the total arrangements of all $8$ letters:

$$8! = 40320$$

Now count the arrangements in which all the vowels come together. Treat $A, U, E$ as one single group. Then the entities are: vowel-group, $D, G, H, T, R$. So total entities become $6$.

$$6! = 720$$

The vowels within the group can be arranged among themselves in:

$$3! = 6$$

Hence, the number of arrangements in which all vowels come together is:

$$6! \times 3! = 720 \times 6 = 4320$$

Therefore, the required number of arrangements in which all vowels do not come together is:

$$8! - 6! \times 3! = 40320 - 4320 = 36000$$

Therefore, the number of such words is $36000$. The correct option is C.

Given: All letters of the word "DAUGHTER" are to be used. Find: The number of arrangements in which the vowels are not all together.

1. Total words using all letters:

$$8! = 40320$$

2. Unwanted cases: all vowels together.

Treat the three vowels $A, U, E$ as one block. Then we arrange $6$ objects:

$$(\text{AUE block}), D, G, H, T, R$$

Number of such arrangements:

$$6! = 720$$

3. The vowels inside the block can be permuted in:

$$3! = 6$$

4. So total unwanted arrangements are:

$$6! \times 3! = 4320$$

5. Subtract from the total:

$$40320 - 4320 = 36000$$

Hence, the required number of words is $36000$.