Question 1191617
Here's how to solve this combination problem:

**a) 4 moderately difficult routines:**

* There are 5 moderately difficult routines.
* The student must choose 4 of them.
* Number of ways = ⁵C₄ = 5! / (4! * 1!) = 5

**b) 4 easy or moderately difficult routines:**

* There are 2 easy + 5 moderately difficult = 7 routines that are either easy or moderately difficult.
* The student must choose 4 of them.
* Number of ways = ⁷C₄ = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35

**c) 2 moderately difficult and 2 difficult routines:**

* There are 5 moderately difficult routines, and the student must choose 2: ⁵C₂ = 5! / (2! * 3!) = 10
* There are 3 difficult routines, and the student must choose 2: ³C₂ = 3! / (2! * 1!) = 3
* Since these choices are independent, we multiply the number of ways for each: 10 * 3 = 30

**d) 1 easy and 3 difficult routines:**

* There are 2 easy routines, and the student must choose 1: ²C₁ = 2! / (1! * 1!) = 2
* There are 3 difficult routines, and the student must choose 3: ³C₃ = 3! / (3! * 0!) = 1
* Since these choices are independent, we multiply the number of ways for each: 2 * 1 = 2