Answer:
There are 720 permutations BUT 180 "different" permutations
There are 6 letters (ignoring at the moment that there are 2Fs and 2Es). Therefore there are 6! ways that you can order any 6 letters.
The first can be any of 6
The second can be any of 5 (one already first)
The third can be any of 4 (two already first and second)
The 4th can be any of 3
The 5th can be either of 2
The 6th is whatever is left ie 1
Therefore there are 6*5*4*3*2*1 = 6! or 720 different permutations of any 6 letters. If the 2 "Fs" and 2 "Es" can be distinguished, ie they are different sizes or different colours then that is the answer you require 720 permutations.
HOWEVER, if they CANNOT be distinguished and one F looks just like the other F, then for each and every one of the 720 permutations calculated above there will be an IDENTICAL permutation where we just swap one F for the other one. E.g. F(1)COEEF(2) LOOKS exactly the same as F(2)COEEF(1). They both look like FCOEEF. Therefore we must divide the 720 permutations by 2 to allow for the duplicated Fs. 720 / 2 = 360
We must similarly divide these permutations by 2 to allow for the duplicated Es
360 / 2 = 180
"Different" permutations in this example = 6! / (2!*2!) = 720 /(2*2) = 720/4 = 180
There are 180 DIFFERENT permutations
