An example might help. Consider the text $probability$. If we thought of all letters as distinct: $p\ r\ o\ b_1\ a\ b_2\ i_1\ l\ i_2\ t\ y$, then there would be $11!$ permutations.
But this overcounts, as not all letters are distinct. In particular, amongst the $11!$ factorial arrangements above, we have both $\eqalign{ &r\ p\ o\ b_1\ a\ b_2\ \color{maroon}{i_1}\ l\ \color{maroon}{i_2}\ t\ y\cr &r\ p\ o\ b_1\ a\ b_2\ \color{maroon}{i_2}\ l\ \color{maroon}{i_i}\ t\ y\cr} $ For any particular arrangement amongst the $11!$ arrangements of the letters, we can find one other (by permuting the $i$'s) that is the same when we regard the $i$'s as indistinguishable.
So we've over counted by a factor of 2. Similarly, because there are two $b$'s, we overcounted by another factor of 2.
So, the correct number of arrangements is $11!\over 2\cdot2$.
A similar argument can be made to establish the formula that Hooked gives in his answer. In particular if one letter was repeated thrice, then if we found the number of arrangements where we thought of all letters as distinct, we would have over counted by a factor of $3!$; since there are $3!$ ways to permute the three repititions of the letter.