In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?

We are given that 2 courses are compulsory out of the 9 available courses,

Thus, a student can choose 3 courses out of the remaining 7 courses.

Number of ways $={ }^{7} C_{3}=\frac{7 !}{3 ! 4 !}=\frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}=35$