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.

• A. $4320,36000$
• B. $4300,36000$
• C. $4200,36000$
• D. $4300,3600$

Answer 1

Answer: (A)

(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,
$\therefore$ 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 $A,U,E$ taken all at a time. Hence, by the multiplication principle, the required number of permutations $=6!\times 3!=4320$.
(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 $=8!-6!\times 3!=6!(7\times 8-6)=2\times 6!(28-3)=50\times 6!=50\times 720=36000$