How many ways can you pick a team of 6 players from a class of 25 students, with no restrictions, if Harry is not on the team, and if Fred and Harry always fight so you can't have both of them on the team?

Here are the solutions to the problems:

Question 5: How many ways can you pick a team of 6 players from a class of 25 students?
This is a combination problem because the order in which players are chosen for a team does not matter. The formula for combinations is $C(n, k) = \frac{n!}{k!(n-k)!}$.

• a) With no restrictions We need to choose 6 players from 25 students.

  $$C(25, 6) = \frac{25!}{6!(25-6)!} = \frac{25!}{6!19!}$$ $$C(25, 6) = \frac{25 \times 24 \times 23 \times 22 \times 21 \times 20}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$$ $$C(25, 6) = 25 \times \frac{24}{6 \times 4} \times 23 \times \frac{22}{2} \times \frac{21}{3} \times \frac{20}{5 \times 1}$$ $$C(25, 6) = 25 \times 1 \times 23 \times 11 \times 7 \times 4$$ $$C(25, 6) = 177100$$

  There are $\boxed{\text{177,100 ways}}$ to pick a team with no restrictions.

• b) If Harry is not on the team If Harry is not on the team, we exclude him from the pool of students. This means we are choosing 6 players from the remaining $25 - 1 = 24$ students.

  $$C(24, 6) = \frac{24!}{6!(24-6)!} = \frac{24!}{6!18!}$$ $$C(24, 6) = \frac{24 \times 23 \times 22 \times 21 \times 20 \times 19}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$$ $$C(24, 6) = \frac{24}{6 \times 4} \times 23 \times \frac{22}{2} \times \frac{21}{3} \times \frac{20}{5} \times 19$$ $$C(24, 6) = 1 \times 23 \times 11 \times 7 \times 4 \times 19$$ $$C(24, 6) = 134596$$

  There are $\boxed{\text{134,596 ways}}$ if Harry is not on the team.

• c) If Fred and Harry always fight so you can't have both of them on the team This means we cannot have both Fred and Harry on the team. We can find this by taking the total number of ways (from part a) and subtracting the number of ways where both Fred and Harry are on the team.

  First, calculate the number of ways where both Fred and Harry are on the team:
  If Fred and Harry are both on the team, then 2 spots are filled. We need to choose $6 - 2 = 4$ more players from the remaining $25 - 2 = 23$ students.

  $$C(23, 4) = \frac{23!}{4!(23-4)!} = \frac{23!}{4!19!}$$ $$C(23, 4) = \frac{23 \times 22 \times 21 \times 20}{4 \times 3 \times 2 \times 1}$$ $$C(23, 4) = 23 \times \frac{22}{2} \times \frac{21}{3} \times \frac{20}{4}$$ $$C(23, 4) = 23 \times 11 \times 7 \times 5$$ $$C(23, 4) = 8855$$

  Now, subtract this from the total number of ways (from part a):
  Number of ways = (Total ways) - (Ways with both Fred and Harry)

  $$177100 - 8855 = 168245$$

  There are $\boxed{\text{168,245 ways}}$ if Fred and Harry cannot both be on the team.