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Find the number of arrangements that can be made of the letters of the word 'DIFFERENT' without changing the place of a vowel?
20180312 20:21:54
My procedure :
The word given is 'DIFFERENT'
The condition that the position of a vowel cannot be changed means that permutations of vowels will not be counted. So I have cancelled $3$ letters 'I' , 'E' and 'E' because they are not part of my condition.
So now we have total $6$ numbers, out of them $2$ are common i.e.,'F ', 'F'.
After calculation, I got the result $360$.
$\dfrac{6!}{2!}= 360$ ; here '!' Indicates factorial
But in my book the result is given $359$, Can anyone help to find where I have made a mistake? Or you can do the full math and show me how you obtained $359$?
Yes it is correct, maybe they have excluded DIFFERENT itself by the counting.

Yes it is correct, maybe they have excluded DIFFERENT itself by the counting.
20180312 22:23:51