This isn't quite correct. Each word has one of the following sets of I's and E's
I, E, II, IE, EE, IIE, IEE, IIEE
In the case of I or E we have \(\displaystyle \dbinom{2}{1}\dbinom{4}{1}3!\)
The first factor selects whether we have I or E. The second factor selects the slot in the word for it.
Then the third factor recognizes that we use the 3 letters not I or E and all permutations of them are valid.
In the case of IIE or IEE we have \(\displaystyle \dbinom{2}{1}\dbinom{4}{1}\dbinom{3}{1}\dbinom{3}{1}\)
The first factor selects whether we have two I's or two E's, the second factor selects the slot for the single I or E,
the third factor selects the letter from the other 3, and the fourth factor selects the slot for the other letter.
I'll let you work up the numbers for IE, II, EE, and IIEE. It's all very similar to above.
Then add up all numbers. You should get 270.