Who says math doesn’t relate to real life? Here is a question that came up in our office just the other day:

*An office comprised of eight employees is planning to have a foosball game. A matchup consists of four players, split into pairs. If any employee can be paired up with any other employee, then how many unique matchups result?*

*(A) **70*

*(B) **210*

*(C) **280*

*(D) **336*

*(E) **420*

I know I said no more combinations problems but try this question, inspired by another real-life scenario. At Magoosh, we have a foosball table along with several avid players. We enjoy playing when we are not busy ‘Magooshing’.

The great thing about this problem is it requires you to think a little bit. You can’t simply use the combinations formula and hope that an answer will just pop out, as if by magic.

**Solution**

Sure there is 8C4 in there somewhere; after all, we have to choose four from eight. But here is where things get a little tricky. If you have Al, Bob, Chris (that’s me), and Dave playing a game, then, according to the conditions in the problem, you can break up these four as follows: Al can play with three different people.

Therefore, for any four players, there are three possible matchups. So how many ways can you choose four players from eight? That’s right, now it’s time to bring in our combinations formula (click here if you want to see a short cut for dealing with these problems). 8C4 = 70. Because we have three possible matchups from those 70 different ways of choosing four players, we get 70 x 3 = 210. Answer B.

**Takeaway**

**An interesting takeaway from this problem – at least in my opinion – is becoming aware of the presence of math in our lives. That’s right numbers and knotty problems – like our foosball conundrum – are all around you. Time to leave a tip at a restaurant? Wondering the number of ways your friends can sit while on a road trip? Need to figure out how many miles per gallon you’re getting on that road trip? (Given that your car doesn’t have one of those newfangled dashboards that does all the work for you).**

**In fact, I like this idea so much, I’m going to come up with more GRE-style problems based on real-life situations. Indeed, feel free to offer any real-life based mathematical conundrums you’ve encountered.**

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Hi Chris,

I am just wondering why I cannot do 8C2 * 6C2 in this case. Please explain. Many thanks.=)

hi chris,

I couldn’t get one thing in the solution. Why did you multiply 3 in the end ( with 8C4)? When we simply do 8C4, aren’t we considering all the possibilities in which we could form the team?

Hi Jay,

Within the four chosen there can be three possible teams. To illustrate: If we have a team consisting of A, B, C, and D we can break these players up into the following:

AB vs CD

AC vs BD

AD vs BC

Hope that helps 🙂

Thanks Josh for commenting.

Chris, please don’t be sorry. This is a good example you have posted.

Great, I am happy people like this example. It is a fun little one with a twist. Not too much math but very easy to get wrong (just like a real GRE question :).

Hi Chris,

Thank you for the problems you have been posting.

For the problem mentioned above, I had a different solution as below:

1. Choose 4 of 8 = 8C4

2. Out of the chosen 4, I have to pair 2 =4C2=6 ways

Eg: (1,2,3,4 =pairs 12,13,14,23,24,34=6 ways in which 4 players have unique match-ups)

Hence 70 x 6 =420

What am I doing wrong here?

The first part 8C4 is correct but can you further explain 70×3=210. I understand what you have written and it makes sense but can that be derived mathematically?

shri, I was confused about the same thing but I think I figured it out:

I think it is because if the pair 12 is chosen, then the second pair must be 34. When you are choosing pairs from a group of 4 people, once you have chosen the first pair, the second pair is automatically chosen. So choosing the pair 12 is the same as choosing 34. So the six ways that you mentioned above should be combined into the 3 ways: (12|34), (13|24), and (14|23).

Sorry shri for never replying – for some reason your comment got lost in the mix.

Josh’s reasoning is solid – basically when you’ve chosen one group, let’s say A and B, the other group has to be C and D. Therefore you have to divide by 2.