Let's solve each question one by one.

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Question 6

How many ways can 5 students be assigned a seat in a classroom if there are 8 seats in a row?

This is a problem of selecting and arranging 5 students in 8 available seats, which requires a combination of selection and permutation.

1. Choose 5 seats out of 8: We can choose 5 seats from the 8 seats in $\binom{8}{5}$ ways.
2. Arrange 5 students in those 5 chosen seats: For each selection of seats, we can arrange 5 students in $5!$ (5 factorial) ways.

So, the total number of ways is: $\binom{8}{5} \times 5! = \frac{8!}{5!(8-5)!} \times 5! = \frac{8!}{3!} = \frac{40,320}{6} = 6,720$

Answer: 6,720

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Question 7

Suzy randomly picks marbles from a bag containing 18 identical marbles. How many possible outcomes are there if she selects 2 marbles?

Since the marbles are identical, any selection of 2 marbles will be indistinguishable from any other selection of 2 marbles. Thus, there is only 1 possible outcome when picking 2 identical marbles.

Answer: None of the above

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Let me know if you'd like further details on either problem or a breakdown of any concepts.

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Related Questions

1. How many ways can 5 students be seated if the row has only 5 seats?
2. How would the answer change if the marbles were unique instead of identical?
3. What would the answer to question 6 be if the row had 10 seats instead of 8?
4. How would the calculation change in question 6 if each student had a specific seat preference?
5. If there were 20 identical marbles and Suzy could pick any number between 1 and 4, how many unique outcomes would there be?

Tip:

When calculating combinations and permutations, remember to distinguish between problems involving identical versus unique items, as this affects how outcomes are counted.