How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Probability in Games | AIME I, 1999 | Question 13

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.

## Probability in Games - AIME I, 1999 Question 13

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 $log_{2}n$

• 10
• 742
• 30
• 11

### Key Concepts

Probability

Theory of equations

Combinations

AIME, 1999 Q13

Course in Probability Theory by Kai Lai Chung .

## Try with Hints

First hint

${40 \choose 2}$=780 pairings with $2^{780}$ outcomes

Second hint

no two teams win the same number of games=40! required probability =$\frac{40!}{2^{780}}$

Final Step

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.