Step 1 :Let's solve the problem in two parts. For part a), we need to choose 15 questions out of 20, order doesn't matter. So we can use the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\) where n is the total number of items, k is the number of items to choose, and '!' denotes factorial.
Step 2 :Substituting the given values into the formula, we get \(C(20, 15) = \frac{20!}{15!(20-15)!} = 15504\).
Step 3 :For part b), we first choose exactly 3 questions from the first 5, and then choose the remaining 12 questions from the remaining 15. So we can use the combination formula twice and multiply the results together.
Step 4 :Substituting the given values into the formula, we get \(C(5, 3) * C(15, 12) = \frac{5!}{3!(5-3)!} * \frac{15!}{12!(15-12)!} = 4550\).
Step 5 :Final Answer: a) The number of ways to select 15 questions out of 20 is \(\boxed{15504}\). b) The number of ways to select 15 questions out of 20 with exactly 3 of them being from the first 5 is \(\boxed{4550}\).