Total ways of arranging the 5 letters will be 5!/2! = 60.
Then you need to subtract the cases where you have the two I's next to each other. In such cases, they'd create one single, indivisible group. So you can safely say that the number of cases in which you have II is the number of possible arrangements of (G), (H), (II) and (J) or possible arrangements of 4 objects, i.e. 4! = 24.
You're left with the answer you provided.

DanaJ wrote:Total ways of arranging the 5 letters will be 5! = 60.
Then you need to subtract the cases where you have the two I's next to each other. In such cases, they'd create one single, indivisible group. So you can safely say that the number of cases in which you have II is the number of possible arrangements of (G), (H), (II) and (J) or possible arrangements of 4 objects, i.e. 4! = 24.
You're left with the answer you provided.

Well total ways of arranging 5 letters with 2 common letters=5!/2!=60 and 4!=24
Answer should be 36

Yeah, you're right. I was thinking it like that, but didn't exactly write it down as such.

Hey guys why are you dividing 5!/2!?
I thought the answer was 5! - 4! = 96. Why will you divide by 2! when the letters are common??
I simply took it as. Since they are five letters, the total number of ways they can be arranged with repetitions are 5! (without repetitions they will be 5!/2 though) so with repetitions the answer is 96 and without it is yeah 36

why the 5 ways to arrange 5 letters 60? shouldn't be 5! only?

You need to divide by 2! because you have two letters that are basically the same: I and I. Say you have I(1) and I(2). There's no difference between HI(1)I(2)GJ ans HI(2)I(1)GJ, since you won't see those numbers when it comes down to it.
5! applies only when you do not have any repeated letters.