Question

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

The Correct Option is A

(i) There are $8$ different letters in the word , 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 $= 8! - 6! \times 3! = 6! (7 \times 8 - 6) = 2 \times 6! (28-3) = 50 \times 6! = 50 \times 720 = 36000$