This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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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) = (n!)/(k!(n-k)!). a) With no restrictions We need to choose 6 players from 25 students. C(25, 6) = (25!)/(6!(25-6)!) = (25!)/(6!19!) C(25, 6) = (25 × 24 × 23 × 22 × 21 × 20)/(6 × 5 × 4 × 3 × 2 × 1) C(25, 6) = 25 × (24)/(6 × 4) × 23 × (22)/(2) × (21)/(3) × (20)/(5 × 1) C(25, 6) = 25 × 1 × 23 × 11 × 7 × 4 C(25, 6) = 177100 There are 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) = (24!)/(6!(24-6)!) = (24!)/(6!18!) C(24, 6) = (24 × 23 × 22 × 21 × 20 × 19)/(6 × 5 × 4 × 3 × 2 × 1) C(24, 6) = (24)/(6 × 4) × 23 × (22)/(2) × (21)/(3) × (20)/(5) × 19 C(24, 6) = 1 × 23 × 11 × 7 × 4 × 19 C(24, 6) = 134596 There are 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) = (23!)/(4!(23-4)!) = (23!)/(4!19!) C(23, 4) = (23 × 22 × 21 × 20)/(4 × 3 × 2 × 1) C(23, 4) = 23 × (22)/(2) × (21)/(3) × (20)/(4) C(23, 4) = 23 × 11 × 7 × 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 168,245 ways if Fred and Harry cannot both be on the team.