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 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
(i) To find the number of permutations where O and A occupy the end places, we fix O and A at the last two positions. The other three letters, M, E, and G, can be arranged in the remaining three positions in 3! = 6 ways.
Therefore, the number of permutations with O and A at the end is 6.
(ii) To find the number of permutations where E is always in the middle, we fix E in the middle position. The other four letters, O, M, G, and A, can be arranged in the remaining four positions in 4! = 24 ways.
Therefore, the number of permutations with E in the middle is 24.
(iii) To find the number of permutations where the vowels occupy odd places, we fix the vowels, O, E, and A, in the odd positions. The consonants, M and G, can be arranged in the even positions in 2! = 2 ways.
Therefore, the number of permutations with vowels in odd places is 2.
(iv) To find the number of permutations where the vowels are never together, we consider the number of ways to arrange the vowels O, E, and A such that they are not together.
We can treat the three vowels as a single unit. The number of ways to arrange this unit along with the consonants M and G is 3! = 6.
However, within the unit of vowels, there are 3! = 6 ways to arrange the vowels themselves. Out of these 6 arrangements, only 2 arrangements have the vowels together (OEA and AEO).
Therefore, the number of permutations where the vowels are not together is 6 - 2 = 4.
To summarize:
(i) Number of permutations with O and A at the end: 6
(ii) Number of permutations with E in the middle: 24
(iii) Number of permutations with vowels in odd places: 2
(iv) Number of permutations where vowels are never together: 4
To find the number of permutations in each case, we need to apply combinatorial principles. Let's tackle each case one by one:
(i) O and A occupying end places:
In this case, we have fixed positions for O and A at the end. The remaining four letters (M, E, G, and A) can be arranged among themselves in 4! = 24 ways.
However, since 'A' appears twice in the word 'OMEGA', we need to divide by 2! (the factorial of the number of times A appears) to account for the repetitions.
So, the total number of permutations is 4! / 2! = 24 / 2 = 12.
(ii) E being always in the middle:
In this case, we fix the position of E in the middle. Now we have five remaining letters to arrange: O, M, G, A, and A.
The number of ways to arrange these five letters is 5!.
However, since 'A' appears twice in the word 'OMEGA', we again need to divide by 2!.
So, the total number of permutations is 5! / 2! = 120 / 2 = 60.
(iii) Vowels occupying odd places:
In this case, there are two vowels in the word 'OMEGA': O and A. We want to place them in the odd positions (1st, 3rd, and 5th places).
First, we arrange the three vowels in these odd positions: O _ A _ A.
The letter M and G can be placed in the remaining two even positions (2nd and 4th places).
The number of ways to arrange the consonants is 2! (as there are two different consonants).
Thus, the total number of permutations is 2! = 2.
(iv) Vowels being never together:
To find the number of permutations where vowels are never together, we will use the principle of inclusion-exclusion.
We first calculate the total number of permutations without any restrictions. In this case, we have six distinct letters, so the total number of permutations is 6! = 720.
Next, we calculate the number of permutations where the vowels are together. We can treat 'AO' as one entity (grouping both vowels together), so we have five entities to arrange: AO, M, G, and E.
The number of ways to arrange these five entities is 5!.
However, within the 'AO' entity, there are two identical letters (A), so we need to divide by 2!.
Thus, the number of permutations with vowels together is 5! / 2! = 120.
Finally, we subtract the number of permutations with vowels together from the total number of permutations to get the number of permutations with vowels not together:
720 - 120 = 600.
To summarize:
(i) O and A occupying end places: 12 permutations.
(ii) E being always in the middle: 60 permutations.
(iii) Vowels occupying odd places: 2 permutations.
(iv) Vowels being never together: 600 permutations.