How many words can be formed from the letters of the word daughter so that D is always next to E but before G?

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How many words can be formed from the letters of the word "d a u g h t e r" so that the vowels never come together?

There are $3$ vowels and $5$ consonants. I first arranged $5$ consonants in five places in $5!$ ways. $6$ gaps are created. Out of these $6$ gaps, I selected $3$ gaps in ${}_6C_3$ ways and then made the vowels permute in those $3$ selected places in $3!$ ways. This leads me to my answer $5!\cdot {}_6C_3 \cdot 3! = 14400$.

The answer given in my textbook is $36000$. Which cases did I miss? What is wrong in my method?

How many words with or without meaning can be formed from the letters of the word, 'DAUGHTER', so thati all vowels occur together?ii all vowels do not occur together?

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Text Solution

Solution : The letters of the word daughter are “d,a,u,g,h,t,e,r”.<br> So, the vowels are ‘a, u, e’ and the consonants are “d,g,h,t,r”.<br> (i)Now, all the vowels should come together, so consider the bundle of vowels as one letter, then total letters will be `6`.<br> So, the number of words formed by these letters will be `6!`<br> but, the vowels can be arranged differently in the bundle, resulting in different words, so we have to consider the arrangements of the `3` vowels.<br> So, the arrangements of vowels will be `3!`<br> Thus, the total number of words formed will be equal to `(6!×3!)=4320`<br> (ii)First arrange `5` consonants in five places in `5!` ways.<br> `6` gaps are created. Out of these `6` gaps, select `3` gaps in `6_(C_3)`​ ways and then make the vowels permute in those `3` selected places in `3!` ways.<br> This leads `5!×6_(C_3)​xx3!`=14400.