Combinations problem: Choosing ways to select $8$ questions out of $12$. In an exam, there are $12$ questions in total. He has to attempt $8$ questions in all. There are two parts: Part A and Part B of the question paper containing $5$ and $7$ questions respectively. How many ways are there to attempt the exam, such that you attempt eight questions and from each part you attempt at least three questions? 
Please don't answer with the three case method. What I want is the error in my method? 
My method: For three questions from each part, we do: $${5 \choose 3}*{7\choose3} $$. But 2 questions are lett, so for that, we multiply the expression by $${6\choose2} $$. Because 6 questions are left to choose from. But this does not give the answer, why?

You are counting certain combinations of questions more than once. This is not so easy to see for your example, thus, let's simplify it to a trivial example: you have to choose at least 5 questions out of 5 of part A (i.e. all of A), and at least 1 out of 7 of part B, and, as before, you have to choose 8 questions in total.
We can trivially see that we indeed have to choose 3 questions out of B to satisfy both conditions, since we have already chosen all of A. Thus, the correct answer would be $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\3\end{matrix}\right)=1\cdot 35=35$. 
Your way of solving this question would however say that there are $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\1\end{matrix}\right)\left(\begin{matrix}6\\2\end{matrix}\right)=1\cdot 7\cdot 15=105$
possibilities of choosing.
Edit: to clarify the problem in this simplified example: assume you choose questions 1,2 and 3 from B. In your calculations there are three possibilities to do so: you choose question 1 because you have to choose at least one from B (Condition 1), and you choose questions 2 and 3 because you have to choose in total 8 (Condition 2). Or you choose question 2 to satisfy Condition 1, and questions 1 and 3 for Condition 2. Finally, you can choose question 3 to satisfy Condition 1, and questions 1 and 2 for Condition 2. In other words, you count each combination 3 times, and, indeed $35\cdot3=105$.
