The SAT will throw two common kinds of “counting” problems your way. I’ll handle one of them in this post. The other kind, well, I’ll get to it when I get to it. 🙂

I like to call this kind of problem a matching problem. It’ll usually involve a bunch of people who all need to shake hands, or a league in which every team needs to play every other team. Or a massage club where everyone has to give everyone else a back massage. I don’t know…whatever. Everyone has to touch everyone at the touching party?

…this is going nowhere. Let’s see an example.

1. Each team in a kickball league plays each other team 4 times during the season. If there are 7 teams in the league, how many games long is the season?

(A) 28
(B) 56
(C) 84
(D) 112
(E) 116

What you’re going to want to do here is draw a diagram.

Arrange the letters A-G (representing the 7 teams) in a large circle. Now draw
lines connecting each letter to each other letter, carefully counting as you draw. (If you try to count after you’re done drawing, you’re going to have a pretty difficult time getting an accurate count.) The best way to go about this is to draw every line that you can that originates at
A, and then do the same for B, etc.

You’ll know you’re done when you have something that resembles a star with all its outer points connected. Like so:

The number of lines you just drew – 21, you awesome counter you – equals the number of games required for each team to play each other once. If each team has to play each other 4 times, multiply 21 by 4 to get the answer: 84! BAM!

Pretty amazing, right? It’s just so…beautiful. No, stop crying. It’s totally inappropriate for you to be crying right now. I know it’s pretty but you need to stop. I refuse to move on until you stop crying.

Of course, if you want to represent the above diagram mathematically, you could say that Team A needs to play all the other teams, so it plays 6 games. Then Team B needs to play all the teams except Team A (since they already played), so that’s 5 more games. Follow that line of reasoning until its end and you get:

6 + 5 + 4 + 3 + 2 + 1 = 21

But I think the star is prettier (and easier to remember).

##### Care to try a few more?

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Jack12 says:

these were pretty easy here is how I did them.

17.Tteam 1 plays against the 6 other teams and each of them four times so tm 1 plays 24 games. Team 2 plays also plays the other 6 teams but since team 1 was already counted it plays 20 games. So now with the pattern of -4 games you just have to add 24+20+16+12+8+4+0 = 84. Really easy once you get it.

16. The exact same concept as above. The first person shakes hands with five others. The second person can only shake hands with four others because he already shook hands with the first. Thus its 5+4+3+2+1 = 15

17. Again same concept. Including pete there are a total of 5 players. 4+3+2+1 = 10. Then there is the final match, 10+1 = 11.

But the star thing was pretty cool too.

MIT/Harvard Tutor says:

Yet again, there’s a much more intuitive and deeper way to solve the problem. Anytime N teams must play each other exactly once, aka anytime N people must shake each other’s hands exactly once (the problem is generally known as the Handshake Problem), there are (N choose 2) games played — you’re just choosing the two teams to play, and you don’t care about order because a Yankees-Red Sox game is the same thing as a Red Sox-Yankees game (in this scenario, where who’s home doesn’t matter…). So the fast, deep, generalizable way to answer your example is 7C2 times 4, 84.

MIT/Harvard Tutor says:

The interesting insight you can generate here by connecting the method I just described to the sum-up-all-the-possibilities method (6+5+4+3+2+1) is that the sum of the first N positive integers is just (N+1 choose 2). That’s because (N+1 choose 2) is just (N+1)!/[2!*(N-1)!]=(N+1)(N)/2=the formula for the sum of an arithmetic sequence with initial term 1 and final term N.

Yes! I love working through this with students who have already done some combinatorics in school and seeing the light bulb above their heads flicker on. 😉

Chris says:

A good trick answer for the example problem would have been “21”. I got to that point on my own and thought that that was the answer. If it had been test day, I would have been doomed!