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  $\binom{12}{3}\;$
 A: 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.  
A: The generating function for the number of $k$ courses such that there at least one course from each category is the following:
$$\Bigg( \binom{3}{1}x + \binom{3}{2}x^2 + \binom{3}{3}x^3 \Bigg) \Bigg(\binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \binom{6}{4}x^4 +\binom{6}{5}x^5 + \binom{6}{6}x^6 \Bigg)^2$$
Thus the number of ways to choose $6$ courses with at least one from each category is the coefficient of $x^6$ in the generating function above.
