The six letters of the word COFFEE are rearranged. In how many of the resulting letter-strings do the two F's stand together AND the two E's also stand together?

Glue the two F's into one block and the two E's into another. The units left to arrange are [FF], [EE], C and O — four distinct units, giving 4! = 24 strings. Nothing further happens inside either block, because the two F's are identical and so are the two E's. Trap $96=24\times 2\times 2$ shuffles each block internally, which produces nothing new when the letters are alike. Trap 180 = 6!/(2! × 2!) is the unrestricted total for COFFEE. Trap 120 glues only the F's and then treats the two E's as though they were different letters, giving 5! instead of the correct 5!/2! = 60 for that half-done case.