Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.
Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find \(log_{2}n\)
Probability
Theory of equations
Combinations
Answer: 742.
AIME, 1999 Q13
Course in Probability Theory by Kai Lai Chung .
\({40 \choose 2}\)=780 pairings with \(2^{780}\) outcomes
no two teams win the same number of games=40! required probability =\(\frac{40!}{2^{780}}\)
the number of powers of 2 in 40!=[\(\frac{40}{2}\)]+[\(\frac{40}{4}\)]+[\(\frac{40}{8}\)]+[\(\frac{40}{16}\)]+[\(\frac{40}{32}\)]=20+10+5+2+1=38 then 780-38=742.
