Here are three questions covering permutations, combinations, and probability distribution, including a table summarizing them:

Topic	Question
Permutations	How many distinct ways can 5 students be seated in 5 chairs in a row?
Combinations	From a deck of 52 cards, how many different ways can you choose 3 cards from the set of 13 spades?
Probability Distribution	A fair 6-sided die is rolled twice. What is the probability distribution for the sum of the two rolls?

1. Permutation Question:

Question: How many distinct ways can 5 students be seated in 5 chairs in a row?

Solution:
This is a permutation problem because the order matters. The number of ways to arrange 5 students in 5 chairs is given by $5!$ (5 factorial):

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Thus, there are 120 distinct seating arrangements.

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2. Combination Question:

Question: From a deck of 52 cards, how many different ways can you choose 3 cards from the set of 13 spades?

Solution:
This is a combination problem since the order of the cards doesn’t matter. The number of ways to choose 3 cards from the 13 spades is given by the combination formula $\binom{n}{r}$, where $n = 13$ and $r = 3$:

$\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$

So, there are 286 ways to choose 3 spades.

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3. Probability Distribution Question:

Question: A fair 6-sided die is rolled twice. What is the probability distribution for the sum of the two rolls?

Solution: We calculate the sum of the two rolls, which can range from 2 (1+1) to 12 (6+6). There are 36 possible outcomes (since $6 \times 6 = 36$). The probabilities for each sum can be calculated by counting how many ways each sum can occur.

Sum	Number of Ways	Probability
2	1 (1+1)	$\frac{1}{36}$
3	2 (1+2, 2+1)	$\frac{2}{36}$
4	3 (1+3, 2+2, 3+1)	$\frac{3}{36}$
5	4 (1+4, 2+3, 3+2, 4+1)	$\frac{4}{36}$
6	5 (1+5, 2+4, 3+3, 4+2, 5+1)	$\frac{5}{36}$
7	6 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)	$\frac{6}{36}$
8	5 (2+6, 3+5, 4+4, 5+3, 6+2)	$\frac{5}{36}$
9	4 (3+6, 4+5, 5+4, 6+3)	$\frac{4}{36}$
10	3 (4+6, 5+5, 6+4)	$\frac{3}{36}$
11	2 (5+6, 6+5)	$\frac{2}{36}$
12	1 (6+6)	$\frac{1}{36}$

So, the probability distribution for the sum of the two rolls is as shown above.

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Would you like more details or explanations for any of these questions? Here are five related questions to consider:

1. How would the seating arrangement change if some students refuse to sit next to each other?
2. What if you choose 5 cards from the 13 spades? How many combinations are possible?
3. What is the expected value of the sum in the probability distribution for two dice?
4. How does the probability distribution change if one of the dice is biased?
5. How do you calculate permutations if some items are repeated?

Tip: Always distinguish between permutations (where order matters) and combinations (where order doesn't) when solving problems.