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# How many possible March Madness combinations are there?

I have been thinking about this for a while, but I can't figure out the math behind it.

There are 65 teams which compete in the NCAA Basketball tournament In March of every year.

I assume that you know what the bracket looks like, but I will go over it a bit:

teams 64 and 65 play the first round to get into the top 64 teams. After this each team plays another, eventually eliminating the opponents until one team remains.

My question is how many different possible combinations of wins/losses can there be?

### 7 Answers

- 1 decade agoFavorite Answer
I heard on sports talk radio last week there are over 9 quintillion brackets. I believe that is 9 with 18 zeros after it.

Also, there is a website of a mathematician who provides the same quote on his website.

In response to truthbetold's... it would be 2^64 if you count the play-in game, otherwise it would be 2^63 which is 9.223 quintillion not counting the play-in game.

- conventionalLv 41 decade ago
its easy, each game can have 1 of 2 possible outcomes. with 64 games, this gives 2^64 or 18 quintillion possibilities... 18,446,744,073,709,551,616.

This is about 18 million trillions.

good luck filling out all those brackets.

if this doesnt seem to make sense, think of a simpler case, how many combination of flipping two coins? its 2^2 = 4. HH, HT, TH, and TT. or 3 coins is 2^3 = 8. HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. you can see how that can be extrapolated to 64 coins, or 64 games in our case, yea?

- Anonymous1 decade ago
Exactly 18,446,744,073,709,551,616

Edit: Just so you guys know, that's not a joke. There are 64 games.. 2 ^ 64 is 18,446,744,073,709,551,616

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- Kyle MLv 61 decade ago
a quick guesstimate would be somehting like 32! plus 16! plus 8! plus 4! plus 2! plus 1 or something.