The number of ways in which we can form a 8 letter word from the letters of the word DAUGHTER such that all vowels are never occur together is
My approach:
As they are 5 consonants(DGHTR) and 3 vowels(AUE)
we can arrange 5 consonants in 5! ways.
$$*D*G*H*T*R*$$
Now 6 empty slots can be filled with either all vowels separately (A,U,E), which can be done in
$\binom{6}{3}\times3!$ ways
or 6 empty slots can be filled with a vowel and 2 vowel occurring together (A,UE or U,AE or E,AU) which can be done in
$\binom{6}{2}\times2!+\binom{6}{2}\times2!+\binom{6}{2}\times2!=3\times\binom{6}{2}\times2!$
Summing up all the possibilities = $5!\times(\binom{6}{3}\times3!+3\times\binom{6}{2}\times2!)=25200$
But answer is given as 36000, What possibility I missed out.
Best Answer
I don't know what possibility you missed out, but I'd think it's easier to calculate how many possibilities there are where all the vowels do appear together, and subtract that from the total $8!=40\,320$.
If all the vowels appear together, then we can think of those three together as one letter, which means there are $6!$ different words we can write. But for each of those there are $3!$ ways of rearranging the vowels within the block. So there are $6!\cdot 3! = 4\,320$ different words where all the vowels appear together.
This means there are $40\,320 - 4\,320 = 36\,000$ words where all the vowels do not appear together.