For a very long time, we’ve had Vocab Weds, a weekly vocab post with an accompanying YouTube video. It’s done quite well, and we have no intention of stopping it. But poor math—it’s gotten no such weekly billing. So to redress this oversight, we are going to inaugurate the weekly math “Brain Twister” question. It will be as tough—if not a tad tougher—than the most difficult question you can expect to see on the GRE.

This week, in keeping with the World Cup spirit, we’ve come up with a fun—but fiendish—question. You should give yourself no more than 3 minutes (this applies to all of the Clemmonsdogpark “Brain Twisters”) with each question, the assumption being that the hardest question on the GRE math should not take you any longer (if it does, skip it).

And if you think you know the answer, comment below. Explanations are very welcome (and the first correct answer always gets bragging rights!) I’ll post the explanation every Thursday.

## Clemmonsdogpark’s First Brain Twister: World Cup Edition

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

The answer is Option E.

Reason: Since we have four groups of 8 each, the number of different ways 4 teams can qualify for the semis is 8 * 8 * 8 * 8 = 2^12. In the semis, take any one team, there are 3 other teams to choose from to make a semi-final match up. (Once one semifinal is fixed, the other gets fixed by default). So there are totally 3 * (2^12) distinct possible semi-final matches.

Hope that is correct.

2^12 matches till semis.

4 teams will make it to semis. There could be a total of 3 possible unique matches.

Ans: 2^12*3

I think its C …as we have to select 4 teams and then each team will play …with rest , keeping in mind …the matches need to be different

It is not clear if you want to count all possibilities so that:

A vs B, C vs D is one option

A vs B, C vs E is another option

As opposed to

A vs B is one option

C vs D is another option

Judging by your answers you meant the former though to me the wording of the question indicates the latter interpretation.

…I guess you got us on this one! 😀

That wasn’t my intention :(. Hopefully somebody gets the answer. Though the explanation post is going up later today, so we’ll see.

Actually, I looked through the comments again and realized that two students below did answer the question correctly 🙂

There are 4 groups of 8 teams.

A semi-final matchup will be winning team from those 4 groups.

Therefore, there are 8^4 or 2^12 possibilities (D).

I am not sure about (E) though.

I could not resolve why there is an extra 3.

Good luck!

You are very close, but there is a little twist at the end that accounts for the 3 (Hint: the set up differs from the World Cup, which, in terms of semifinals possible matchups, would be 2^12.

Chris,

I see no restriction in forming the 8 groups consisting of 4 teams each,so there should be 32C4 ways of forming each group, any of which can make it to the semifinal. I do not see this to be one of the answer choices, so I must be missing something.

– Sriram.

Hi Sriram,

There is actually a restriction: teams in the same bracket (group of 8 teams) can’t face off against each other in the semifinals because only one of those eight can make it to one of the four semifinals spots.

See if that helps :). If not, the explanation is going to appear later today.

Hi Chris,

Thanks, yes, I did figure out the answer and responded yesterday. What is you see is my older post. Some gremlins at work I guess, this is the second time my comments have vanished.

That’s weird Sriram :(. Hopefully, it’s not happening with everyone (though very few mention that their comments vanish).

Well, for what its worth,here’s what I wrote:

Looking at the responses, I see that I completely misread the statement “smaller grouping of eight teams, one of which will play in the semi-finals, which comprise the final four teams” as 8 teams(or groups) each containing 4 teams, one of which would be playing in the semi-final. Now, if we’re ‘starting’ with 4 teams(or groups),each containing 8 teams already chosen – this is what the clue suggests as different from the current FIFA groupings,I presume – then we’re looking at 8C1 ways of picking one team from each group – thus resulting in (8C1)^4 ways as already commented. Finally, if “many distinct semi-final matchups” means the count of the semifinalist teams likely, the computation ends here. Else,if we’re required to compute the total count of matches that can result out of these chosen semi-finalists, then we perform an additional step of 4!/2!2!2! and multiply by 8^4. So its D or E, as correctly pointed out by a few responses.

Option D?

Close, but think about the possible arrangements of semifinalists 🙂

I think it would be B. 32!/4!

Reason: The 32 teams can play each other in 32! ways. Since the teams would be split in 4 distinct groups, hence the 4!.

Divya,

Remember that teams in the same bracket can’t play each other in the semifinals.

Hopefully that helps :). If not, the explanation will be up later today.

Good luck!

I am screwed up. May be c but not clear at all. 🙁

It’s a tough question. In fact, nobody has yet gotten it :(. We’ll have the explanation up later today.

Out of 8 teams in each team one team will be in semis, so there are 8C1 ways to do it. Since (8C1)^4 = 2^12 ways. Now these four teams will play each other, So we have 4!/2!2!2! = 3 ways to do it. So answer is 3*2^12 ie E.

Wow, I didn’t see this comment till now! But yes, you are the first (and only) person to get it right :).

Awesome!

There are 4 groups of 8.1 from each group will play the semis.There are 8*8*8*8 different ways this can happen or 2^12 ways This final four can play themselves in 3 ways resulting in (d)2^12 * 3 distinct matches

Actually, it looks like you got it too! However, you put (D), so I’m only realizing it now–after looking a little bit more closely–that you actually meant (E).

Awesome job!

Is it B?

That’s a trap answer. The teams in the same bracket can’t actually play each other in the semifinal so that limits the number of distinct semifinal matchups.

Take another shot at it 🙂