Since the order of students is not important, this is a combinatorics problem. We are asked to find how many ways five students can be selected from a group of 33 students.

The formula for combinations is:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Where:

• $n$ is the total number of students (33),
• $r$ is the number of students to choose (5),
• "!" denotes factorial.

Now we can calculate:

$\binom{33}{5} = \frac{33!}{5!(33-5)!} = \frac{33!}{5!28!}$

I'll compute this.There are 237,336 different ways for five students to form a group for the activity.

Would you like further explanation or details?

Here are 5 related questions to consider:

1. How many ways can you select 3 students from the 33 students?
2. What if the order of selection mattered? How many different groups could you form then?
3. How many ways can you form a group of 5 students from a class of 40 students?
4. What if 2 specific students must be included in the group? How many ways can you select the remaining students?
5. How many ways can you form two groups of 5 from the class of 33 students?

Tip: When order doesn't matter, use combinations; when it does, use permutations!