Asked by lovemaan5500 | 2nd Feb, 2019, 08:29: PM
i) No two vowels are together
According to the condition we need to arrange consonants first. The number of consonants are 5.
Hence, these can be arranged by 5! = 120 ways
_ C _ C _ C _ C _ C _
If the vowels can be placed on the empty space by 6P3 ways i. e. 6 × 5 × 4 = 120 ways
Total number of words = 120 × 120 = 14400
ii) The vowels can occupy odd places
Odd places are 1st, 3rd, 5th, 7th
These can be done in 4P3 = 4 × 3 × 2 = 24 ways
The remaining 5 consonants can be occupied in 5! ways i. e. 120
Total number of ways = 24 × 120 = 2880
Answered by Sneha shidid | 4th Feb, 2019, 10:05: AM
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