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$$ \binom{12}6 = \frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 924. $$ Sometimes it's hard to talk students out of computing both the numerator and the denominator in this expression and then dividing.

I can think of three reasons for this:

  • It takes at least some effort to learn that one can simplify things like this by canceling.
  • Elementary school pupils are told $3\times4$ means "multiply $3$ by $4$", i.e. "$\times$" is a verb in the imperative mood. Even in elementary school, I reminded myself every time I heard this that $3\times4$ is a noun, but I've never seen any evidence of anyone else doing that.
  • Calculators are anesthetics. Students approach math problems with great anxiety, and, confronted with an expression like the one above, frantically reach for their calculators the way a drowning person reaches for anything that floats. A calculator WILL get the student out of the terrible predicament that is a math problem and calculators are INFALLIBLE. If $8/3=2.666666\ldots$ and a calculator says $2.666\times51=135.966$ then that is EXACT and INFALLIBLE. Whoever says the answer is $136$ is rounding it. If $\pi=3.14$ and you find that $3.14^3\cong30.959$, and you want to be more accurate, just add more of the digits that the calculator shows you when you find $3.14^3=30.959144$, and if the exam question says you are to give an EXACT answer, that means give all the digits that your calculator will show you. Calculators COMPLETELY OBVIATE ALL NEED TO THINK ABOUT ANY QUESTIONS THAT THIS PRESENT PARAGRAPH MIGHT SUGGEST. Anyone who doubts that is a lunatic.

In simplifying rational functions, plainly one should cancel first, but with rational numbers, it's not always clear to student what the advantage is, if any. Are there examples not involving algebra, but only arithmetic, that are as cogent to newbies as are examples of simplifying rational functions?

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  • $\begingroup$ Why don't you tell us how you really feel about calculators? $\endgroup$ – Kevin Carlson Oct 1 '12 at 18:06
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    $\begingroup$ @KevinCarlson : I use calculators and I like calculators, but I don't bow down in idolatrous prayer to them. $\endgroup$ – Michael Hardy Oct 1 '12 at 18:14
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    $\begingroup$ I've found another good rule of thumb is, whenever student laziness is suggested as a cause (like here: it takes effort to learn cancellation and they haven't put any effort in) then one should also consider if fear of making a mistake is at fault. No doubt, many such students have been duly punished for some cancelling mistakes. $\endgroup$ – rschwieb Oct 1 '12 at 20:45
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    $\begingroup$ It's often better NOT to cancel first. Canceling is typically ad hoc and introduces a chance for errors. Also, if I'm doing such a calculation as part of a search for a general solution to something, I want to see the denominator in its natural form. Easier to look up in OEIS and so forth. $\endgroup$ – I. J. Kennedy Oct 1 '12 at 20:55
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    $\begingroup$ Once I asked on a test for the students to write the prime factorization of $2^{14}3^55^37^{12}(13)^2(23)^3-2^73^55^67^{12}(13)^2(23)^3$. And every one of them tried multiplying the two terms out! $\endgroup$ – Lubin Oct 2 '12 at 2:54
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The trouble is that students at that level look at a formula as an instruction.

Here’s an example of the kind you’re requesting, and I think it shows the wrongness of “formula=instruction”. Evaluate this compound fraction: $$ \frac{\frac{1}{2}-\frac16}{\frac34+\frac5{12}}\,. $$ Students will invariably subtract the two fractions upstairs, add the two fractions downstairs, and then divide one fraction by the other. But this is not the efficient way to do it. Remind them that any time you have a fraction, it is legal to multiply top and bottom by the same number. So we choose to multiply top and bottom of the big fraction by 12, getting $$ \frac{6-2}{9+5}=\frac{4}{14}=2/7\,. $$ No addition of fractions, no division of fractions, in spite of the appearance of the original form.

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  • $\begingroup$ But please tell them why they can multiply above and below! $\endgroup$ – JP McCarthy Nov 14 '13 at 21:53
  • $\begingroup$ My opinion, @JpMcCarthy, is that with proper prior training, they shouldn’t have needed to be told why. $\endgroup$ – Lubin Nov 14 '13 at 22:29
  • $\begingroup$ What if they don't have proper prior training? What if they think they can add above and below? $\endgroup$ – JP McCarthy Nov 14 '13 at 22:45
  • $\begingroup$ This is getting too far afield, but they can be disabused of the notion that you can add above and below by noting that $1/2\ne2/3$. $\endgroup$ – Lubin Nov 15 '13 at 16:21
  • $\begingroup$ So they can multiply above and below but not add. What about the poor soul who says what is the 'rule' again and gets it the wrong way around? Better they know why they can multiply above and below so there isn't a rule but rather a simple idea. $\endgroup$ – JP McCarthy Nov 15 '13 at 16:53
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I'm not really sure if I'm addressing your question the way you want but here's a simpler version of the combinations question that makes it pretty obvious:

$$ \frac{6^{1432}}{6^{1430}}=\text{?} $$

I think the combinations example is rather good. It's really easy to sell by saying: "if you cancel first, you will multiply less numbers and smaller numbers."

The only ones that don't comply are the students that are hopelessly entrenched in the mentality that everything you are trying to teach them is meant to complicate what they already know. It's probably also intertwined with a deep fear of making cancellation mistakes.

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You could always make up example that are above the calculator's memory, assuming they don't do their calculations on computer. Ask them to tell you what $ 148\choose 72 $ is or anything involving numbers above $70!$

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