Let's work through each of these problems step-by-step from your document:

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Q1: How many hockey teams of 6 players can be formed from 14 players?

To find the number of ways to form a hockey team of 6 players from a group of 14 players, we use the combination formula:

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

where $n = 14$ and $r = 6$.

Calculating:

$C(14, 6) = \frac{14!}{6!(14-6)!} = \frac{14!}{6! \times 8!}$

Simplifying this:

$C(14, 6) = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 3003$

Answer: There are 3,003 ways to form the hockey team.

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Q2: Selecting a 4-person bobsled team with one member from 9 football players and the other 3 from a group of 8 athletes.

1. Choose 1 person from the group of 9 football players:

$C(9, 1) = 9$

2. Choose 3 people from the group of 8 athletes:

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

3. Total number of ways to form the team:

$9 \times 56 = 504$

Answer: There are 504 ways to form the bobsled team.

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Q3: An airline is hiring 5 flight attendants from 8 applicants.

We use the combination formula to find the number of ways to select 5 attendants from 8 applicants:

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

Answer: There are 56 ways to hire the flight attendants.

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Q4: Forming a committee of 5 men and 6 women from a group of 10 men and 12 women.

1. Select 5 men from 10:

$C(10, 5) = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$

2. Select 6 women from 12:

$C(12, 6) = \frac{12!}{6!(12-6)!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$

3. Total ways to form the committee:

$252 \times 924 = 232,848$

Answer: There are 232,848 ways to form the committee.

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Q5: Selecting the first, second, and third place winners from 8 students.

This is a permutation problem where order matters. The formula is:

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

where $n = 8$ and $r = 3$.

Calculating:

$P(8, 3) = \frac{8!}{(8-3)!} = \frac{8 \times 7 \times 6}{1} = 336$

Answer: There are 336 ways to select the winners.

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Q6: Selecting 3 pieces of fruit from 5 different pieces by 3 people.

The number of ways for 3 people to choose 1 piece each from 5 different pieces is a permutation since order matters:

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3}{1} = 60$

Answer: There are 60 ways for the 3 people to make their choices.

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Would you like more details on any of these solutions? Here are five related questions for deeper exploration:

1. What if the order did not matter in the fruit selection problem?
2. How do combinations change if the number of players or applicants changes?
3. What if some positions in the bobsled team required specific skills?
4. How do permutation and combination calculations change when repetitions are allowed?
5. Can you explain the difference between combinations and permutations with examples?

Tip: When determining whether to use a combination or permutation, remember that combinations are used when the order doesn't matter, and permutations are used when the order does matter.