To solve this problem, we need to determine in how many ways a student can choose a program of 5 courses out of 9 available courses, given that 2 specific courses are compulsory for every student.
- First, identify the compulsory courses. Since 2 courses are compulsory, we start by including these courses in the selection automatically.
- We have a pool of 9 courses in total. After accounting for the 2 compulsory courses, we are left with: \(9 - 2 = 7\) courses.
- Out of these remaining 7 courses, the student needs to choose the remaining 3 courses to complete the set of 5 courses.
- The number of ways to choose 3 courses from the 7 remaining courses is given by the combination formula: \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), where \(n\) is the total number of courses to choose from, and \(r\) is the number of courses to choose.
- Substitute \(n = 7\) and \(r = 3\) into the formula to find the number of ways: \(\binom{7}{3} = \frac{7!}{3! \times (7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\).
- Therefore, there are 35 ways to choose 3 courses from the 7 available courses after considering the 2 compulsory courses.
Thus, the correct answer is 35.