In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A,B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is_________.

To solve this problem, we need to consider all possible combinations of questions a student can attempt from each section while adhering to the condition of attempting at least 4 questions from each section.

1. The student can attempt 4 questions from section A, 5 from section B, and 6 from section C. The number of ways to do this is C(8,4)*C(6,5)*C(6,6), where C(n,r) is the number of combinations of n items taken r at a time.

2. The student can attempt 5 questions from section A, 4 from section B, and 6 from section C. The number of ways to do this is C(8,5)*C(6,4)*C(6,6).

3. The student can attempt 5 questions from section A, 5 from section B, and 5 from section C. The number of ways to do this is C(8,5)*C(6,5)*C(6,5).

4. The student can attempt 5 questions from section A, 6 from section B, and 4 from section C. The number of ways to do this is C(8,5)*C(6,6)*C(6,4).

5. The student can attempt 6 questions from section A, 4 from section B, and 5 from section C. The number of ways to do this is C(8,6)*C(6,4)*C(6,5).

6. The student can attempt 6 questions from section A, 5 from section B, and 4 from section C. The number of ways to do this is C(8,6)*C(6,5)*C(6,4).

7. The student can attempt 7 questions from section A, 4 from section B, and 4 from section C. The number of ways to do this is C(8,7)*C(6,4)*C(6,4).

Adding up all these possibilities gives the total number of ways a student can select 15 questions.