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The number of ways of arranging all the ...

The number of ways of arranging all the letters on a circle such that no two vowels are together.

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There are `21` consonants and `5` vowels.
First we will arrange `21` consonants.
They can be be arranged in `(21-1)! = 20!` ways.
Now, we can place 5 vowels in between consonants.
It can be done in `C(21,5)` ways. Also, these `5` vowels can be arranged in `5!` ways.
So, vowels can be placed between consonants in `C(21,5)*5!` ways.
So, required number of arrangements ` = C(21,5)*20!*5!`
Also, if we consider clockwise and anti-clockwise arrangements same, then the required number of arrangements `= (C(21,5)*20!*5!)/2`
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