Asked by "ThoughtWaves "
Find the number of permutations that can be had form the letters of the word ‘OMEGA’
(i) O and A occuping end places. (ii) E being always in the middle.
(iii) Vowels occuping odd places. (iv) Vowels being never together.
(i) O and A occuping end places. (ii) E being always in the middle.
(iii) Vowels occuping odd places. (iv) Vowels being never together.
Answers
Answered by
GPT 3.5
(i) O and A occupying end places:
Since O and A must occupy the end places, we have 4 remaining letters to place in between them. Therefore, there are 4! = 24 permutations.
(ii) E being always in the middle:
Since E must be in the middle, we have 5 remaining letters to place on either side of E. Therefore, there are 5! = 120 permutations.
(iii) Vowels occupying odd places:
There are 3 vowels in the word 'OMEGA' (O, E, and A). We can choose one of the vowels for the first odd place (3 choices), then one of the remaining two vowels for the third odd place (2 choices), and finally have one letter (G or M) remaining for the fifth odd place (2 choices). Therefore, there are 3 * 2 * 2 = 12 permutations.
(iv) Vowels being never together:
To ensure that the vowels are never together, we can place the vowels in the consonant positions. There are 3 consonants in the word 'OMEGA' (M, G, and M). We can choose one of the 3 consonants for the first position (3 choices), then one of the remaining 2 consonants for the third position (2 choices), leaving one consonant for the fifth position (1 choice). Finally, there are 3! = 6 permutations for the remaining three vowels. Therefore, there are 3 * 2 * 1 * 6 = 36 permutations.
In summary:
(i) O and A occupying end places: 24 permutations
(ii) E being always in the middle: 120 permutations
(iii) Vowels occupying odd places: 12 permutations
(iv) Vowels being never together: 36 permutations
Since O and A must occupy the end places, we have 4 remaining letters to place in between them. Therefore, there are 4! = 24 permutations.
(ii) E being always in the middle:
Since E must be in the middle, we have 5 remaining letters to place on either side of E. Therefore, there are 5! = 120 permutations.
(iii) Vowels occupying odd places:
There are 3 vowels in the word 'OMEGA' (O, E, and A). We can choose one of the vowels for the first odd place (3 choices), then one of the remaining two vowels for the third odd place (2 choices), and finally have one letter (G or M) remaining for the fifth odd place (2 choices). Therefore, there are 3 * 2 * 2 = 12 permutations.
(iv) Vowels being never together:
To ensure that the vowels are never together, we can place the vowels in the consonant positions. There are 3 consonants in the word 'OMEGA' (M, G, and M). We can choose one of the 3 consonants for the first position (3 choices), then one of the remaining 2 consonants for the third position (2 choices), leaving one consonant for the fifth position (1 choice). Finally, there are 3! = 6 permutations for the remaining three vowels. Therefore, there are 3 * 2 * 1 * 6 = 36 permutations.
In summary:
(i) O and A occupying end places: 24 permutations
(ii) E being always in the middle: 120 permutations
(iii) Vowels occupying odd places: 12 permutations
(iv) Vowels being never together: 36 permutations