Friends Different Teams 0219B7

1. **Problem statement:** We want to find the percentage probability that each of the 4 friends is on a different team when 25 players (4 friends + 21 others) are divided into 5 teams of 5 players each. 2. **Recall the probability formula:** $$P = \frac{14400 \times \binom{21}{4} \binom{17}{4} \binom{13}{4} \binom{9}{4}}{\binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5}}$$ 3. **Calculate each combination:** - $\binom{25}{5} = 53130$ - $\binom{20}{5} = 15504$ - $\binom{15}{5} = 3003$ - $\binom{10}{5} = 252$ - $\binom{5}{5} = 1$ - $\binom{21}{4} = 5985$ - $\binom{17}{4} = 2380$ - $\binom{13}{4} = 715$ - $\binom{9}{4} = 126$ 4. **Calculate numerator:** $$14400 \times 5985 \times 2380 \times 715 \times 126$$ 5. **Calculate denominator:** $$53130 \times 15504 \times 3003 \times 252 \times 1$$ 6. **Calculate numerator value:** $$14400 \times 5985 = 86184000$$ $$86184000 \times 2380 = 205237920000$$ $$205237920000 \times 715 = 146750402800000$$ $$146750402800000 \times 126 = 18490551152800000$$ 7. **Calculate denominator value:** $$53130 \times 15504 = 823993152$$ $$823993152 \times 3003 = 2474704623456$$ $$2474704623456 \times 252 = 623630764709312$$ 8. **Calculate probability:** $$P = \frac{18490551152800000}{623630764709312} \approx 29.65$$ 9. **Convert to percentage:** $$29.65 \times 100\% = 2965\%$$ This is impossible as probability cannot exceed 100%, so let's check the calculation carefully. 10. **Recalculate denominator multiplication:** $$53130 \times 15504 = 823993152$$ $$823993152 \times 3003 = 2474704623456$$ $$2474704623456 \times 252 = 623630764709312$$ 11. **Recalculate numerator multiplication:** $$14400 \times 5985 = 86184000$$ $$86184000 \times 2380 = 205237920000$$ $$205237920000 \times 715 = 146750402800000$$ $$146750402800000 \times 126 = 18490551152800000$$ 12. **Divide numerator by denominator:** $$\frac{18490551152800000}{623630764709312} \approx 29.65$$ 13. **Error found:** The probability cannot be greater than 1. The mistake is that the teams are unlabeled, so the total number of ways should be divided by $5! = 120$ in the denominator, which was already accounted for in the formula. 14. **Recalculate denominator with division by 120:** $$\text{Denominator} = \frac{53130 \times 15504 \times 3003 \times 252 \times 1}{120}$$ $$= \frac{623630764709312}{120} = 5196923039244.27$$ 15. **Calculate final probability:** $$P = \frac{18490551152800000}{5196923039244.27} \approx 3.56$$ Still greater than 1, so let's check the approach. 16. **Alternative approach:** Probability that each friend is on a different team is: $$P = \frac{5}{5} \times \frac{4}{24} \times \frac{3}{23} \times \frac{2}{22}$$ This is the probability that the second friend is on a different team than the first, the third friend on a different team than the first two, and so on. 17. **Calculate this probability:** $$P = 1 \times \frac{4}{24} \times \frac{3}{23} \times \frac{2}{22} = \frac{4 \times 3 \times 2}{24 \times 23 \times 22} = \frac{24}{12144} \approx 0.001977$$ 18. **Convert to percentage:** $$0.001977 \times 100\% = 0.1977\%$$ **Final answer:** The probability that each of the 4 friends plays on a different team is approximately $0.198\%$.