Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that:

(i) all vowels occur together      

(ii) all vowels do not occur together.

(i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely A, U, and E. Since the vowels have to occur together, we can for the time being assume them as a single object (AUE). This single object together with the 5 remaining letters (D, G, H, T, R) will be counted as 6 objects.

Number of permutations of these 6 objects taken all at a time = 6!

Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time.

Hence, by the multiplication principle, the required number of permutations = 6! × 3! = 720 × 6 = 4320.

(ii) If we have to count those permutations in which all vowels are not together, we first find all possible arrangements of 8 letters taken all at a time, which can be done in 8! ways.

Then, we subtract from this number the arrangements in which the vowels are always together.

Therefore, the required number = 8! − 6! × 3!

= 40320 − 4320 = 36000.

Hence,

(i) When all vowels occur together → 4320 arrangements
(ii) When vowels do not occur together → 36000 arrangements