Math, asked by somyag9065, 8 months ago

write all permutation of examination if the vowels are never together

Answers

Answered by ahervandan39
0

Answer:

There are 5 letters in the word OMEGA

(i) When O and A occupying end places M,E,G,(OA)

The first 3 letters can be arranged in 3! ways and the letters O,A can be arranged in 2! ways.

∴ Total number of such ways = 6×2=12 ways

(ii) When E is fixed in the middle, the remaining 4 words can be arranged in 4! ways.

∴ Total number of such words = 4!= 24 ways

(iii) 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.

∴ Total number of such words = 3!×2!= 12 ways

(iv) Total number of words = 5! = 120

Combine the vowels into one bracket as (OEA) and treating them as one letter we have three letters (OEA),M,G and these can be arranged in 3! ways and three vowels can be arranged in 3! ways.

∴ Number of ways when the vowels come together = 3!×3! = 36 ways

Hence, the number of ways when vowels being never together = 120−36=84 ways

The required answer =

6

84−12−24−12

=6

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