How to Count

The other day I saw a math question disguised as a baseball trivia question. Here it is:

How many states don’t have major league baseball teams?

Let’s see: there’s Alaska, Arkansas,… Sure, it might be hard to list them all, but why am I calling this a math question?

Well, it doesn’t ask us to list all the states without baseball teams, it asks us to count them. Of course you can count things directly, by listing every one, but that’s not always as easy as it might seem. Maybe you can list all 50 states off the top of your head, and keep track as you go along of which ones don’t have teams, but I’m pretty sure I’ll overlook a few states. (I thought I could organize the states alphabetically, but I ended up giving up once I thought I got through the A’s, and I forgot Alabama!)

So how do you count things indirectly, without listing them? For a start, let’s reframe the question:

How many states DO have major league baseball teams?

If we can answer one question, we can answer the other: if, say, 20 states out of 50 have teams, then 30 don’t. But doesn’t the question feel a little easier when you ask it this second way? Pause here with me for just a moment: why is that?

One reason is that relatively few states have teams, and the ones that do are likely to be the better known ones, so if you were going to try to count by listing, listing the states that have teams is probably easier than listing the ones that don’t. But the real reason the alternate formulation helps is that you don’t have to count by listing — at least not by listing states. You could count by listing teams.

The Red Sox play in Massachusetts — that’s one state. The Yankees play in New York — that’s a second. The Giants play in California — a third. The A’s play in California too, but we already counted that. And so on.

We can make this process a little more organized if we use the structure of the baseball leagues. There are 30 major league baseball teams and they are currently divided evenly into two leagues: 15 in the American League, 15 in the National. Each league has 3 divisions — East, Central, and West — and each division has 5 teams. In other words: 30 teams broken up into 6 divisions of 5.

Doesn’t it feel a lot easier to go through 6 divisions of 5 than to go through 50 states? Let’s do it. I write this off the top of my head, in real time:

AL East: Boston Red Sox (MA, 1), New York Yankees (NY, 2), Baltimore Orioles (MD, 3), Toronto Blue Jays (Canada, not a state), Tampa Bay Rays (FL,4)

AL Central: Kansas City Royals (MO, 5), Detroit Tigers (MI, 6), Cleveland Indians (OH, 7), Minnesota Twins (MN, 8), Chicago White Sox (IL, 9)

AL West: Oakland A’s (CA, 10), Houston Astros (TX, 11), Texas Rangers (TX, repeat state), California Angels (CA, repeat state), Seattle Mariners (WA, 12)

NL East: Washington Nationals (DC, not a state), New York Mets (NY, repeat state), Philadelphia Phillies (PA, 13), Miami Marlins (FL, repeat state), Atlanta Braves (GA, 14)

NL Central: St. Louis Cardinals (MO, repeat state), Pittsburgh Pirates (PA, repeat state), Milwaukee Brewers (WI, 15), Cincinnati Reds (OH, repeat state), Chicago Cubs (IL, repeat state)

NL West: Arizona Diamondbacks (AZ, 16), Colorado Rockies (CO, 17), San Diego Padres (CA, repeat state), Los Angeles Dodgers (CA, repeat state), San Francisco Giants (CA, repeat state)

And there you have it: 17 distinct states with teams, so 33 states without. And while this problem isn’t winning anybody the Fields Medal, it does illustrate two very important principles of counting, and math in general:

1. Find and use correspondences. When we asked which states have teams, we set up an implicit correspondence between states and teams. A way to make that correspondence more explicit is to reframe the question yet again, this time in terms of team-state pairs:

How many pairs (S, T) are there, where S is a state, T is a team that plays in that state, and no state is repeated more than once?

This might sound needlessly complicated, but math people actually like to talk this way! (Remember the definition of relations and functions the first time you saw it? Your eyes probably glazed over; mine sure did.) We use this language because it brings to the surface the duality inherent in the set-up: states and teams are paired. When you have pairs, you get to choose how to enumerate them: over the first entry, or over the second. And in this case, the second is the way to go, because…

2. More structure is better. The set of states seems sort of amorphous. You can try to break it up into regions (New England, Mid-Atlantic, Midwest,…), but it’s not totally clear how to do it. Whereas the set of baseball teams has a very clear structure: six by five. I lied in one place when I told you I was listing baseball teams in real time. When I got to the NL Central, I put down three of the five teams, and then spaced on what the other two were. But I knew there had to be five, and I knew about where they should be geographically. I remembered the other two within a minute.

Counting has a rich and noble history. Also a fancier name: combinatorics. And while the subject, perhaps like much of math, might seem like a bag of tricks when you first encounter it, it has some clear guiding principles. Look for structures, and try to transform your problem so you can make use of those structures. These principles are at work all over, so keep an eye out for them!


Ee-ther/Ai-ther: Calling the Whole Thing Off at the Science Museum

The papers (the Boston Globe, Time, and others) were abuzz yesterday about a supposed error in a math exhibit at the Boston Museum of Science. Most of the interest in the story came from the fact that the issue — described as a minus sign instead of a plus sign in a formula for the golden ratio — was pointed out by a 15 year-old. Frustratingly, none of the articles I saw included any actual math, though if you’re familiar enough with the golden ratio, you might guess even from the very brief description above that that the fuss was probably about a difference of convention rather than any kind of serious mistake.

And, right on schedule, today the Globe reports that the exhibit is correct after all. So what’s going on?

Let’s start with what we mean by “golden ratio.” I’ve posted about it before, in the context of ratios of successive Fibonacci numbers, which have the golden ratio as their limit. Let’s start with a picture:220px-SimilarGoldenRectangles.svgIn this picture, the small rectangle (with one side length having length a and the other length b) and the big rectangle (with one side having length a+b and the other having length a) are supposed to be similar, meaning that the ratios of their sides are the same. In other words, if you write the length of the longer side on top, a/b = (a+b)/a. You could also put the length of the shorter side on top and get an equivalent equation: b/a = a/(a+b). Either way, dividing through by top and bottom, we get:

a² = b(a+b),


a² − ab = 0.

This equation has lots of pairs of solutions (a,b). You could find them using the quadratic formula, in one of two ways. If you treat a as the variable, you can solve for it in terms of b:

a = (b ± √b² + 4b²  ) / 2 = b·(1 ± √5 ) / 2.

But the equation is pretty symmetrical, and you can also solve for b in terms of a:

b = (−a ± √a² + 4a²  ) / 2 = a·(1 ± √5 ) / 2.

We need to pare down our solutions just a bit. Knowing that a and b are both lengths of rectangle sides, we should make sure they are both positive. 1 − √5 and 1 − √5 are not positive, so we throw them out, leaving us with

a =  b·(1 + √5 ) / 2   and   b = a·(1 + √5 ) / 2.

Once we know this, it’s easy to talk about ratios of sides. The ratio of the longer side to the shorter side is a/b. Taking the equation a =  b·(1 + √5 ) / 2 and dividing both sides by b, we see that a/b = (1 + √5 ) / 2 = 1.61803… And the ratio of the shorter side to the longer side is b/a, which by similar logic is just (1 + √5 ) / 2 = a/b 1 = 0.61803… (We can also deduce b/a = a/b 1 directly from the initial equation a/b = (a+b)/a, because  (a+b)/a  is just 1 + b/a.)

Pictorially, if the square in our initial picture is 1 × 1 (a = 1), then b = (1 + √5 ) / 2 = 0.61803… (the short side of the small rectangle), and a + b = (1 + √5 ) / 2 = 1.61803… (the long side of the big rectangle).

So what is the golden ratio? Well, which ratio do you want — long side to short side or short to long? Do you say tom-ay-to or tom-ah-to? Which of the two we call golden is unimportant; what matters is that the picture, and all the math around the ratio, are the same either way. Which should we take as the golden ratio? Ee-ther! Or maybe ai-ther!

We think of math as being about deduction and absolute right answers, but it is also full of decisions and conventions. Sometimes the decisions make a difference: we decide to make .9999… equal to 1 (by deciding on certain rules for doing math with infinite sums), and this has consequences across the subject (decimal representations are no longer unique). But sometimes the decisions are only conventions, just a way of fixing language or notation and no more, and don’t matter very much.

We do, however, need to keep track of what conventions we’re using. The 15-year old in the news stories probably learned that the golden ratio is (1 + √5 ) / 2, which is the more common formulation. Then, at the Science Museum, he saw this (photo from the latest Globe article):

image1(10)AIt looked wrong; he was sure it should say (5 + 1) / 2, not (5 − 1) / 2. But read the fine print: the short side divided by the long side. That ratio is indeed (5 − 1) / 2, as the display claims. The Science Museum just happened to frame their display in terms of the opposite ratio from the one the student learned. There’s nothing wrong with that, but we need to be aware that which version of the ratio we use is mathematical convention for us to choose, not mathematical fact set in stone.