The problem you're dealing with is a combinations problem, where the order of selection doesn't matter. You are selecting 5 questions from a list of 8, and the formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

where:

• $n$ is the total number of items (8 study questions),
• $k$ is the number of items to choose (5 questions).

Using this formula, we calculate:

$C(8, 5) = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!}$

Now, simplifying:

$C(8, 5) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$

So, there are 56 different ways for the teacher to choose 5 questions from the list of 8.

Would you like further explanation or details on how the combination formula works?

Here are 5 related questions:

1. What if the teacher wants to choose 6 questions instead of 5?
2. How does this problem change if the order of the questions mattered (permutations)?
3. What if the teacher could select any number of questions from the list?
4. Can this formula be used to calculate other real-life selection problems, like choosing committee members?
5. How would you calculate this if there were 10 questions instead of 8?

Tip: The combination formula can always be used when the order doesn't matter in the selection process.