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 teams win the same number of games is m/n, where m and n are relatively prime positive integers. find log2n

There are (40/2) = 780 total pairings of teams, and thus 2⁷⁸⁰ possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games.
Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total, each team corresponds uniquely with some k, with 0 ≤ k ≤ 39, where k represents the number of games the team won.
With this in mind, we see that there are a total of 40! outcomes in which no two teams win the same number of games.
Further, note that these are all the valid combinations, as the team with 1 win must beat the team with 0 wins, the team with 2 wins must beat the teams with 1 and 0 wins, and so on; thus, this uniquely defines a combination.
The desired probability is thus 40!/2⁷⁸⁰ . We wish to simplify this into the form m/n , where m and n are relatively prime. The only necessary step is to factor out all the powers of 2 from 40!, the remaining number is clearly relatively prime to all powers of 2.

The number of powers of 2 in 40! is

= (40/2) + (40/4) + (40/8) + (40/16) + (40/32)

= 20 + 10 + 5 + 2 + 1

= 38.

780-38 = 742.

Hence, the probability that no two teams win the same number of games is 742.

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