Question

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Topic :- Permutations

Q - All the letters of the word ' EAMCOT ' are arranged in different possible ways . Find the number of arrangements in which no two vowels are adjacent to each other .

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Answer 1

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Vowels in the word 'EAMCOT' =  A,E,O
So, Number of vowels = 3

Now, Arrangement of vowels = 3! =1 x 2 x 3 = 6

Number of arrangement of other letters

= 4! = 1 x 2 x 3 x 4 = 24

Hence, Total number of arrangements = 3! x 4! =6 x 24 => 144




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Answer 2

There Are three consonants ( M, C , T ) and three vowels ( A , E , O ) are in the word EAMCOT.
Since No two vowels have to be together so the possible outcomes are in 4 ways and can be arranged in 4^P3 ways .... And 3 consonants can be arranged in 3 ways....
so 3! × 4!
=> 3 ! = 3× 2× 1= 6
=> 4! = 4× 3 × 2× 1 = 24
So , 3! × 4! = 6 × 24
= 144
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