Question

Find the number of permutations that can be held from the letters of word omega such that all the vowels are neber together

Answer 1

Answer

Total permutations = 3! x 2!= 12 ways

Explanation

Let the given word is 'OMEGA'

The word have 5 letters can be arranged in 5! ways.

But here the question have a condition such that all the vowels are never  together

Here the word consists of 3 vowels and 2 consonants.

So the three vowels O, E and A can only be arranged in odd places and consonants are placed in between vowels.

O, E and A can be placed in 1st, 3rd and 5th places in 3! ways

And consonants M and G places in 2nd and 4th places in 2! ways

Therefore total permutations = 3! x 2!= 12 ways

Answer 2

Given:

Omega

To find:

The number of permutations that can be held from the letters of word omega such that all the vowels are never together

Solution:

Omega = 5 letters

3 vowels

2 consonants

And so,

By permutation,

5 ! ways to arrange the letters

As the vowels should not be together,

They are arranged in alternative places,

Hence,

We get,

3 ! ways

The two consonants can be arranged in,

2 ! ways

Therefore,

Total permutations = 3 !* 2 !

12 ways

Hence, there are 12 ways the letters of word omega such that all the vowels are never together.