Where does understanding math break down?

I received my fall teaching assignment this week.  Next Monday, I stand up in front of 80 or so nervous students who have to pass this class in order to graduate, and I start telling them about math they've probably never seen before.

Most of them already have mental blocks in place telling them that math is difficult.  Many of them struggled in their high school math classes as they moved from arithmetic to algebra and geometry.  Instead of introspecting to figure out what they were doing wrong, they decided it was too "difficult" for them.  So they enter my class with trepidation, and they're already setting themselves up to fail it.

To facilitate learning, one of my tasks is to convince my students that they actually can learn math, that it's not too difficult.  In fact, I contend that the sort of math I'm teaching is, in principle, not difficult at all.  That's what I want to ruminate about in this post.

In this type of math class, we have three basic threads interacting.  First, we have definitions: I write down symbols, I tell you what they mean, I tell you why they mean this, and I give you some simple examples.  Second, we have techniques: I tell you how and why in this class of situations, we use these symbols, and in that class of situations, we use those symbols.  Third, we have problems: I give you a problem, you figure out which class of situations it falls in, and then you use the symbols and their meaning to tell me something about the problem.

Let me give you an example.

The definition. A binomial coefficient is this symbol: for any two positive integers \(n\) and \(k\) with \(n > k\),
\[ C(n,k) = \frac{n!}{k!(n-k)!}.\]
Here \(n! = n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\), the product of all the numbers from \(1\) to \(n\).  In fact, we can write down
\[C(n,k) = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)(k-2)\cdots 3\cdot 2\cdot 1}.\]
That is, we take \(k\) slots, start with \(n\) and continue on down, multiplying as we go, then divide that product by \(k!\).

What it means.  A binomial coefficient tells you how many unordered combinations of \(k\) things you can draw from \(n\) things.  That is, it is the number of \(k\)-element subsets in an \(n\)-element set. This is because the number of ways of picking \(k\) things in order is counted by \(\frac{n!}{k!}\) and the number of ways of ordering those \(k\) things is \(k!\), so to see how many ways there are of picking \(k\) things without regard to order, we divide the number of total ways of picking \(k\) things in order by the number of ways of picking the same \(k\) things in different orders.

Example.  To see this in action, check out
\[ C(5,3) = \frac{5\cdot 4\cdot 3}{3\cdot 2\cdot 1} = 10.\]

The general class of situations.  You use a binomial coefficient any time you want to count the number of ways of choosing \(k\) things from \(n\) things without regard to order.

A particular example.  Say you have five different sandwiches and you grab three of them.  How many different outcomes does this experiment have?  Well, since you're just grabbing them, you're obviously not making a point of checking what order you're taking them.  So you're just choosing \(3\) things from \(5\) things without regard to order.  How many ways are there of doing this?  \(C(5,3) = 10.\)

Where could the process of understanding break down?  You might have just skimmed this because you've already decided you won't ever understand math.  If that's the case, I'd encourage you to go back and read it again.  It's not as bad as you think it is.

Perhaps your eyes glossed over at the symbol \(C(n,k)\).  If that's the case, remember: I told you what the symbol means.  I don't expect you to remember it right away, because it's unfamiliar and new -- but remember that its definition is right there!  Any time you need to remember it, you can just go back and double-check what it means.

Maybe you didn't quite catch why binomial coefficients count unordered combinations.  If so, that's all right; that's a dense paragraph and in the interest of brevity (and accuracy - the whys get short shrift in the classroom) I left out some small steps.  But if you read it a few times, I don't doubt that you'll come to understand.

(I recognize that if this is new to you, you won't catch all of it right away.  I can do this off the top of my head because I've taught this material for several years.  But if you take enough time to practice, you'll come to grasp the idea, just as if you shoot enough baskets, you'll be able to hit a free throw.)

So where else could a student's understanding break down?  There's always the possibility that the student is blowing off the class (because they don't think they're able to do it?).  I don't have too much control over that.

Perhaps students' understanding fails not for any single process, but when many new, related ideas are thrown at them very quickly.  Part of learning mathematics is building up heuristics which indicate which tool to use in a given problem.  If students do not focus sufficiently on distinguishing related ideas from each other, then their heuristics will confuse related but distinct situations and the intuition they develop will not conform to reality.

For example, related to the idea of binomial coefficients measuring unordered combinations is permutation coefficients measuring ordered combinations.  The number of ways of drawing \(k\) items from \(n\) items, counted in order, is \(P(n,k) = \frac{n!}{(n-k)!} = n(n-1)(n-2)\cdots (n-k+1).\)  One way we help students remember is, if there are \(k\) items, you draw \(k\) slots. There are \(n\) ways of filling the first slot, \(n-1\) ways of filling the second slot, and so on.  Then you multiply them all together.  This tool is used to count the number of ordered \(k\)-tuples that can be created from an \(n\)-element set.

If the problem is the same as above --- five sandwiches, you grab three --- a student might reason this way: There are three sandwiches, so I draw three slots.  There are five ways of filling the first slot, four ways of filling the second slot, and three ways of filling the third slot, so there must be sixty ways of drawing three sandwiches from five sandwiches.

Where did the student's reasoning go wrong?  Take a minute and figure it out.

The student's intuition is misshapen: He did not consider order.  The student has internalized that, in any problem that involves drawing items from a larger set, the number of ways of drawing items can be found by drawing a slot for each item, then counting the number of items to be included in each slot.

In principle, (this kind of) math isn't difficult.  It's no harder than learning a foreign language.  You just have to remember the meanings of new symbols, how to use them, and which situations call for using those symbols.  A computer could do it (and anything a computer can do is not difficult, just tedious).

Perhaps failing students just move too fast, gulping instead of chewing, so that their intuitions malfunction?  Or they just don't put in the time and practice necessary for remembering these new, foreign symbols?  Or is there some fundamental limitation that I just don't see?



Edit:
Addendum -- One friend asked me, "Where does 'binomial coefficient' come from?"  A binomial is the sum of two algebraic terms: \(a + b\).  If you take an integer power of a binomial, you get the following polynomial:
\[(a+b)^n = \sum_{k = 0}^n C(n,k)a^kb^{n-k}\]
\[  = a^n + na^{n-1}b + \frac{n(n-1)}{2}a^{n-2}b^2 + \cdots + \frac{n(n-1)(n-2)\cdots 3\cdot 2}{(n-2)(n-4)\cdots 3\cdot 2\cdot 1}a^2b^{n-2} + nab^{n-1} + b^n.\]
So the "binomial coefficients" are coefficients (numbers by which variable expressions are multiplied) in the polynomial you get by raising a binomial to a power.