How many words can be formed with the letters of the word 'DAUGHTER', when
(i) there is no restriction?
(ii) all vowels are together?
(iii) Words start from A?
(iv) words start from A and end with R?

Explanation

No. of letters in the word 'DAUGHTER' =8
(i) If there is no restriction,
No. of permutations to fill 8 places with given 8 letters `= P_(8) = 8! = 40320`.
(ii) In the word 'DAUGHTER'. Vowels = A,U,E and Consonants = D,G,T,H,R
`rArr` No. of vowels and consonants are 3 and 5 respectively.
`:'` A,U,E are always together
`:.` Treating these letters as one letter, the letters are `5 +1 = 6`
Now the number of arrangement of 6 letters at 6 places `= .^(6)P_(6) = 6! = 720` and Number of arrangement of letters A,U,E at 3 places `= .^(3)P_(3) = 3! =6`.
`:.` Required arrangements `= 720 xx 6 = 4320`
(iii) For the words starting with A
A will be at first place and no of arrangements of remaining 7 letters at remaining 7 places
`=.^(7)P_(7) = 7! = 5040`.
(iv) For the words starting with A and ending with R,
No. of arrangements of remaining 6 letters at remaining 6 places
`=.^(6)P_(6) = 6! = 720`.