Think of it this way: start with all the letters a different color, so there are $12!$ different and distinguishable arrangements. Now, paint the $o$'s the same color. Since there are $2!=2$ different (but now indistinguishable) ways that we can arrange the $o$'s, then this cuts the number of distinguishable permutations in half. Likewise when we paint the $t$'s the same color. Then we paint the $i$'s the same color, so since there are $3!=6$ different (but now indistinguishable) ways that we can arrange the $i$'s, then we have cut the number of distinguishable permutations by a factor of $6$. All told, then, there are $$\frac{12!}{2!2!3!}$$ distinguishable permutations.