| 1. |
How many different possible permutations can be made from the word ‘BULLET’ such that the vowels are never together? |
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Answer: Option D Explanation: The word ‘BULLET’ contains 6 letters of which 1 letter occurs twice = 6! / 2! = 360 No. of permutations possible with vowels always together = 5! * 2! / 2! = 120 No. of permutations possible with vowels never together = 360-120 = 240. |
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