Question

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

Answer 1

We can see that the word $OMEGA$ has $5$ letters. There are $3$ vowels $O,E,A$ and $2$ consonants that are $M,G$. Use the conditions to find the number of ways to arrange the letters according to the given conditions. The total number of words that can be formed is $5!$.

Complete step-by-step answer:
(i) The number of permutations with $O$ and $A$ occupying end places be . So, it can be arranged as $MEG\underline {OA}$.
As $OA$ are fixed, the starting three letters $MEG$ can be arranged in $3!$ ways.
The fixed letters $OA$ can be arranged at their fixed places in $2!$ ways.
The number of permutations with $O$ and $A$ occupying end places is:
$\Rightarrow {\text{k}} = 3! \times 2!$
$\Rightarrow {\text{k}} = 6 \times 2$
$\Rightarrow {\text{k}} = 12$

(ii) The number of permutations with $E$ being always in the middle be . So, it can be arranged as $OM\underline E GA$.
As the letter $A$ is fixed at the middle place, the letters $O,M,G,A$ can be arranged in $4!$ ways. So,
$\Rightarrow {\text{m}} = 4!$
$\Rightarrow {\text{m}} = 4 \times 3 \times 2$
$\Rightarrow {\text{m}} = 24$

(iii) The number of permutations with Vowels occupying odd places be . There are three vowels and three odd places (1st, 2nd, 3rd) in this arrangement.
The vowels $O,E,A$ can be arranged in $3!$ ways.
The consonants $M,G$ can be arranged in $2!$ ways. So,
$\Rightarrow {\text{n}} = 3! \times 2!$
$\Rightarrow {\text{n}} = 6 \times 2$
$\Rightarrow {\text{n}} = 12$

(iv) The number of permutations with Vowels never together be . The total number of words that can be formed is $5!$.
Let us consider that all the vowels come together that is $\left( {O,E,A} \right),M,G$. The number of permutations when vowels come together is $3! \times 3!$.
Now, the number of permutations with Vowels being never together can be calculated as:
$\Rightarrow {\text{o}} = 5! - 3! \times 3!$
$\Rightarrow {\text{o}} = 120 - 36$
$\Rightarrow {\text{o}} = 84$
Now, the required value of $o - m - n - k/6$ is:
$\Rightarrow o - m - n - k/6 = 84 - 24 - 12 - \left( {12/2} \right)$
$\Rightarrow o - m - n - k/6 = 48 - 6$
$\Rightarrow o - m - n - k/6 = 42$
Therefore, the value of $o - m - n - k/6$ is $42$.

Note: To find the number of permutations when some letters can never be arranged together, then find the number of permutations when letters can be arranged together and subtract it from the total number of permutations.