There are 12 questions on an exam, and each student must answer only 8 questions including at least 4 of the first 5 questions. How many different combinations of questions could a student choose to answer?
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Split this into 2 cases. Case 1: Student answers 4 of the first 5 questions. This can be done in 5!/4!1! = 5 ways. There are 4 more questions which need to be answered, from a total of 7 questions. So the student can answer 4 of these questions in 7!/4!3! = 7.6.5.4/4.3.2=35 ways. So we have a total of 5.35 possible ways of completing the exam.
Case 2: Student answers all 5 of the first 5 questions. This can clearly be done in only one way. Then we have 3 remaining questions from a total of 7. Which gives us 35 possible ways of answering any three. Totalling 35.1=35
So in total we have 35*5+35=210 possible ways.