Determing the number of possible March Madness brackets

Question

Is there a simple combinatorial explanation to derive the total number of march madness brackets?

Would it be $2*(2^{16}*2^{8}*2^{4}*2^{2}*2)^{2}$ where the final squared takes into account both halves of the bracket?

Answer 1

Since I don't know anything about NCAA's March Madness, I'll assume that you're talking about a 64-team single-elimination tournament.

In the first round, there are $32$ games, each with $2$ possible outcomes. This provides for a total of $2^{32}$ possibilities.

In the next round, there are only $16$ games, again each with $2$ possible outcomes, for a total of $2^{16}$ possibilities.

You keep going, eliminating half of the players each round and taking half as long as the previous round to play the games out, until the finals, at which point there is $1$ game with $2$ possible outcomes.

This is a total of $2^{32} \times 2^{16} \times 2^8 \times 2^4 \times 2^2 \times 2^1$ possible outcomes.

Your answer of $2*(2^{16}*2^{8}*2^{4}*2^{2}*2)^{2}$ is equivalent to this, which means that it's correct. It's a rather roundabout way of coming to the same answer (you don't need to compare each bracket separately), but they are equivalent.