Find the number of arrangements that can be made of the letters of the word 'DIFFERENT' without changing the place of a vowel?

2018-03-12 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.

    2018-03-12 22:23:51