All-different win totals in a 32-team round robin

Thirty-two teams play a round robin, so every pair of teams meets exactly once, and each game is a fair coin flip independent of all others. The probability that no two teams finish with the same number of wins can be written $\dfrac{a!}{2^b}$ with integers $a,b$ and $b$ as large as possible. Find $a+b$.

$528$. There are $\binom{32}2=496$ games. Distinct win counts must be $0,1,\dots,31$, which forces a strict ranking where each team beats exactly those below it; each of the $32!$ orderings gives one outcome. So the probability is $\tfrac{32!}{2^{496}}$ and $a+b=528$.