Question

The number of different 8 letter arrangement that can be made from the letters of word. DAUGHTER so that
(i) all vowels occur together are ...A....
(ii) all vowels do not occur together are ...B...
Here, A and B refer to

• A. 4320,36000
• B. 3600,36000
• C. 4320,30000
• D. None of these

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, we can for the time being, assume them as single object (aue). This single object together with 5 remaining letters (objects) will be counted as a 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
$=8!-6!\times 3!=6!(7\times 8-6)$
$=2\times 6!(28-3)$
$=50\times 6!=50\times 720$
$=36000$