## Question

Thirty-two teams have made the second round of a tournament. In that round, each team will play exactly one team. The winner—and there will always be one winner and one loser—will go on to the next round of sixteen teams, in which this process will continue until there are two teams left to play the final game. Assuming each of the thirty-two teams will be part of a smaller grouping of eight teams, one of which will play in the semi-finals, which comprise the final four teams, how many distinct semi-final matchups can result?

*(Hint: the above differs in one significant way from the actual World Cup groupings).*

- 32×31
- 32!/4!
- 32x31x30x29
- 2^12
- 3 x 2^12

## Answer and Explanation

Brain Twister #1 really seemed to stump people. So far nobody has answered correctly—which wasn’t my intention. Hopefully, there will be a few correct answers by the time this is posted.

Okay, so let’s talk about the wrong answers. I’m going to start with (B) since nobody chose (A), which is incorrect. (B) assumes that the answer relies on the use of combinations. (C), too, takes the same approach but uses the permutation formula.

Both are wrong here, however. See, there’s a twist here. The teams are broken up into four different brackets. It is impossible for teams within the same bracket to play each other in the semifinals because at some point before the semifinals one would have to eliminate the other.

To illustrate, imagine the U.S. and Brazil have to play each other in the first round. There is no way that both of them can emerge victorious from the group of eight, which comprise each bracket. One of them will lose. Therefore, we have to think of the problem in terms of how many different teams can emerge from each bracket. Since each bracket has eight, the answer is 8. The fact that the semifinals are made up of one team from each bracket can be translated as follows: 8 x 8 x 8 x 8.

Notice, we are using the fundamental counting principle here, which tells us to multiply distinct possibilities for each group. 8 is simplified to 2^3 so we end up getting (2^3)^4 = 2 ^12.

However, (D) is not the answer. To many people’s credit, they put (D), which is very close. However, there is yet one more twist. Unlike, the World Cup, the question here does not say which of the four brackets are going to square off against each other. To make this a little less abstract, imagine that the winner of bracket A will play the winner of bracket B, and the winner of bracket C will play the winner of bracket D. This is exactly how the World Cup is set up. However, this question never specifically tells us how the winners of each bracket will be matched up. Therefore, if we take the four teams from the semifinals, let’s call them A, B, C, and D, the semifinals could look like the following:

1. A plays B, and C plays D

2. A plays C, and B plays D

3. A plays D, and B plays C

Since we get three possibilities for semifinal matchups for any group of four teams that reach the semifinals, the answer is 3 x (2^12), or answer (E).

I was getting 2^12*6 too for using 4C2 in the last step. The reason 4C2 is not right is because that formula gives you the #matches possible between 4 teams (AB, AC, AD, CB, CD, BD) BUT #matches ≠ #semifinals, because a semifinal is comprised of 2 matches played simultaneously in two separate stadiums, and so #semifinals = #matches*1/2.

That was indeed a tough problem! 🙂

Hi Albert,

Yes, I think this might be the toughest so far out of all the challenge questions 🙂

Nice explanation.. I was using 4C2 and getting answer as 6..so my final answer was 6×2^12.. I understand why it is 3 now 🙂 thanks

Yep, a lot of people got tricked — not an easy question :).