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

Question

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

Smart Achievers · Accepted Answer

(1)

(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 assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be ${}^{6}P_{6}$ = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U and E taken all at a time. Hence, by the multiplication principle, the required number of permutations is

= 6! × 3! = 4320.

(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangements of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number the number of permutations in which the vowels are always together.

Therefore, the required number

= 8! − 6! × 3!

= 6!(7 × 8 − 6) = 2 × 6!(28 − 3)

= 50 × 6! = 50 × 720 = 36000.

Solution

Q.
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

1 4320, 36000

2 4300, 36000

3 4200, 36000

4 4300, 3600

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