1. The problem is to select three questions from each difficulty level: easy, medium, and hard. 2. This is a combinatorial selection problem where we choose 3 items from each category. 3. The formula for combinations is $$C(n, k) = \frac{n!}{k!(n-k)!}$$ where $n$ is the total number of items and $k$ is the number of items to choose. 4. Assuming there are at least 3 questions in each difficulty, the number of ways to choose 3 from each is: $$C(easy, 3) \times C(medium, 3) \times C(hard, 3)$$ 5. Without specific numbers for total questions in each category, the exact count cannot be computed. 6. If you provide the total number of questions in each difficulty, I can calculate the exact number of ways to select three from each.