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.