<s>[INST] Selecting at least 4 questions of 12 and at least one of the first 3 questions must be taken? 
There are 12 questions. Students must take at least 4 questions and at least one of the first 3 questions is taken too.

How many configurations are there?
My attempt
\begin{align}
\text{answer} &= \sum_{i=1}^3\left({3\choose i}\sum_{j=4-i}^9{9 \choose j}\right)\\
 &= 3415
\end{align}
Is it correct?
 [/INST] Yes that is correct answer. Another way to approach it -
If there are $12$ questions, we have $2^{12}$ ways to select any number of questions including no selection $\big[{12 \choose 0} + {12 \choose 1} + ...+ {12 \choose 12}\big]$.
Out of that ${12 \choose 0} + {12 \choose 1} + {12 \choose 2} + {12 \choose 3} = 299$ are selections with less than $4$ questions.
Similarly if we do not select any of the first $3$ questions, there are $2^9$ ways to select questions and out of that ${9 \choose 0} + {9 \choose 1} + {9 \choose 2} + {9 \choose 3} = 130$ are selections with less than $4$ questions.
So the solution we are interested in $ = 2^{12} - 299 - (2^9 - 130) = 3415$.
</s>