A group of 4 students is selected from a group of 17 students to take part in a class in cell biology. Complete parts (a) and (b) below.
(a) In how many ways can this be done?

There are $◻$ possible ways.
(b) In how many ways will the group who will not take part be chosen?

There are $◻$ possible ways.

Step 1 ：The problem is asking for the number of ways to select a group of 4 students from a group of 17 students. This is a combination problem, as the order of selection does not matter. The formula for combinations is given by: $C(n,k)=\frac{n!}{k!(n-k)!}$ where n is the total number of items, k is the number of items to choose, and '!' denotes factorial.

Step 2 ：For part (a), n = 17 (the total number of students) and k = 4 (the number of students to select for the class).

Step 3 ：For part (b), the group who will not take part will be the remaining students after the 4 students have been selected. This means we are choosing 17 - 4 = 13 students from the group of 17. So, for this part, n = 17 and k = 13.

Step 4 ：Calculating these values, we find that there are 2380 ways to select 4 students from a group of 17, and the same number of ways to select the remaining 13 students.

Step 5 ：This makes sense, as choosing 4 students to participate is equivalent to choosing 13 students not to participate.

Step 6 ：Final Answer: There are $\boxed{{\displaystyle 2380}}$ possible ways to select 4 students from a group of 17 for the class, and there are also $\boxed{{\displaystyle 2380}}$ possible ways to select the group of 13 students who will not take part in the class.