A class has 33 students. In how many different ways can five students form a group for an​ activity? (Assume the order of the students is not​ important.)
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Part 1
There are 
  
enter your response here different ways that the five students can form a group for an activity.
​(Type a whole​ number.)

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!

A class has 33 students. In how many different ways can five students form a group for an activity? (Assume the order of the students is not important.)

Find out how to calculate the number of ways to form a group of 5 students from a class of 33. This problem covers essential combinatorics concepts and uses the combination formula C(n, r) = n! / (r!(n-r)!). Ideal for high school students in Grades 9-12.

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