Distinct permutations of the word "toffee"

We know that the number of permutations of some given string of length $n$ is $n!$, however, we need to take into account the number of repeated permutations, we do this by counting the number of permutations of the repeated letters (in this case $F$ and $E$).

Therefore, we have:

$$\frac{6!}{2!^{2}}=180$$

Hope this helps!

Here is another way to think about it. You have the word TOFFEE and six blanks

- - - - - -

you want to place the letters in. Begin with the E. Choose two of the six blanks and pop the Es in. This can be done ${6\choose 2}$ ways. Now four blanks remain; place the Fs in these. There are ${4\choose 2}$ ways to do this. Two blanks remain for T and O; there are two ways to do this. So you get in toto, $$2{6\choose 2}{4\choose 2} = 2\cdot 15 \cdot 6 =180$$ ways to permute TOFFEE.

The term "distinct permutations" takes into account that the word TOFFEE has two F's and two E's. This means that if we simply swap the two F's that the permutation is considered the same. You have to take this into account when doing the calculations for this problem.