How many words with or without meaning can be formed from the letters of the word daughter if vowels being always together?

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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?

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.