Question

The number of permutations that can be made from the letters of the word $OMEGA$
$\left(i\right)$ with $O$ and $A$ occupying end places be "k"
$\left(ii\right)$with $E$ being always in the middle be "m"
$\left(iii\right)$with Vowels occupying odd places be "n"
$\left(iv\right)$with Vowels being never together be "o"
Find $o-m-n-k/6$ ?

Answer 1

There are $5$ letters in the word $OMEGA$
$\left(i\right)$ When $O$ and $A$ occupying end places $M,E,G,\left(OA\right)$
The first 3 letters can be arranged in $3!$ ways and the letters $O,A$ can be arranged in $2!$ ways.
$\therefore$ Total number of such ways = $6\times 2=12ways$
$\left(ii\right)$ When $E$ is fixed in the middle, the remaining 4 words can be arranged in $4!$ ways.
$\therefore$ Total number of such words = $4!=24ways$
$\left(iii\right)$ Three vowels $(O,E,A)$ can be arranged in the odd places in $3!$ ways and the two consonants can be arranged in $2!$ ways.
$\therefore$ Total number of such words = $3!\times 2!=12ways$
$\left(iv\right)$ Total number of words = $5!=120$
Combine the vowels into one bracket as $\left(OEA\right)$ and treating them as one letter we have three letters $\left(OEA\right),M,G$ and these can be arranged in $3!$ ways and three vowels can be arranged in $3!$ ways.
$\therefore$ Number of ways when the vowels come together = $3!\times 3!=36ways$
Hence, the number of ways when vowels being never together = $120-36=84ways$
The required answer $=\frac{84-12-24-12}{6}=6$