Number of ways to chose 6 courses out of 15 (different) courses We got 3 English courses, 6 Chinese courses, 6 Spanish courses.
Each course is different.
In how many was can we choose 6 courses such that we must chose atleast 1 course from each topic (English, Chinese, Spanish)
If possible, I would like to see a solution without the inclusion - exclusion principle.
edit : I got to $\frac{6*6*3}{3!}*\binom{12}{3}\;$ where $\frac{6*6*3}{3!}$ represents picking a course from each topic (we must choose atleast 1 from each topic), then dividing by the number of permutations. then we are left with 3 courses to choose from 12, so I multiplied by  $\binom{12}{3}\;$

There are some mistakes in your calculation.  Fist, there is no reason to divide by $3!$.   When you choose an English course, a Chinese course, and a Spanish course, the order matters.  You haven't counted anything $6$ times.  Second you have some double counting.  Choosing English 1 and one of the first three courses, and then choosing English 2 as one of the remaining three course is the same as choosing English 2 as one of the first three, and then choosing English 1 as one of the last three.
If you want to do this without inclusion-exclusion (although that's the method I would recommend) note that you can either choose two courses in each language in $${3\choose2}{6\choose2}{6\choose2}$$ ways, or choose two English courses, one in one of the other languages and three of the third in $$2{3\choose2}{6\choose1}{6\choose3}$$ ways, or one English course, two in one of the otherLanguages and three of the third in $$2{3\choose1}{6\choose2}{6\choose3}$$ ways or one English course, one in one of the other languages and four of the third in $$2{3\choose1}{6\choose1}{6\choose4}$$ ways or three English courses, two in one of the other languages and one of the third in $$2{3\choose3}{6\choose2}{6\choose1}$$ ways, if I haven't missed any.  
