Friends Different Teams 7C7Ffb

1. **Problem statement:** We have 4 friends and 21 other boys, making a total of $4 + 21 = 25$ players. They are divided randomly into 5 teams of 5 players each. We want to find the probability that each of the 4 friends is on a different team. 2. **Total number of ways to divide 25 players into 5 teams of 5:** The total number of ways to split 25 players into 5 teams of 5 (where teams are unlabeled) is given by the multinomial coefficient divided by the factorial of the number of teams to account for indistinguishable teams: $$\frac{\binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5}}{5!}$$ 3. **Number of favorable ways (each friend on a different team):** - First, assign each of the 4 friends to different teams. Since teams are unlabeled, we can think of choosing 4 different teams for the 4 friends. - The 4 friends must be placed in 4 different teams, so choose 4 teams out of 5 for them: $\binom{5}{4} = 5$ ways. - Assign each friend to one of these 4 teams: $4! = 24$ ways. - Now, each of these 4 teams already has 1 friend, so each needs 4 more players. - The remaining players are $25 - 4 = 21$ boys. - We need to fill the 4 teams with 4 players each and the 5th team with 5 players. - Number of ways to choose players for these teams: - Choose 4 players for the first team: $\binom{21}{4}$ - Choose 4 players for the second team: $\binom{17}{4}$ - Choose 4 players for the third team: $\binom{13}{4}$ - Choose 4 players for the fourth team: $\binom{9}{4}$ - The remaining 5 players go to the fifth team automatically. - Since teams are unlabeled, divide by $1!$ for the last team. - Total favorable ways: $$5 \times 24 \times \binom{21}{4} \binom{17}{4} \binom{13}{4} \binom{9}{4}$$ 4. **Calculate the probability:** $$P = \frac{5 \times 24 \times \binom{21}{4} \binom{17}{4} \binom{13}{4} \binom{9}{4}}{\frac{\binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5}}{5!}}$$ 5. **Simplify the denominator:** $$\frac{\binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5}}{5!} = \frac{\binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5}}{120}$$ 6. **Rewrite the probability:** $$P = \frac{5 \times 24 \times \binom{21}{4} \binom{17}{4} \binom{13}{4} \binom{9}{4} \times 120}{\binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5}}$$ 7. **Calculate the constants:** $$5 \times 24 \times 120 = 14400$$ 8. **Final probability:** $$P = \frac{14400 \times \binom{21}{4} \binom{17}{4} \binom{13}{4} \binom{9}{4}}{\binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5}}$$ This fraction gives the exact probability that each of the 4 friends is on a different team when 25 players are divided into 5 teams of 5 players each.