Problem-Solving Without Mimicking
Peter Liljedahl has become a watchword in the math education world. He has imbued this world with the mantra that if students are not thinking, they are not learning. Thinking apparently occurs when students are not sitting down, but standing up at “vertical non-permanent surfaces” (sometimes called “white-boards”), and working in groups solving problems for which, generally and in most cases, they have little or no prior knowledge to know how to solve. When solving problems for which they do have such prior knowledge (i.e., they can connect it to similar problems and principles), Liljedahl calls this “mimicking”, which has become enjoined to the previously mentioned mantra as “mimicking isn’t thinking.”
Actually, the Liljedahl phenomenon has served to cast in concrete the reform math tsunami that has overtaken many K-8 classrooms. His approach expands the reform approach to include students standing up and working at white boards. In general, the central ideas of reform math—i.e., collaboration and having students solve new and novel problems for which they have little or no background knowledge—tends or seems to prevail. (Sorry for the “tends or seems to”, but I was trying for a nuanced tone here.)
At his website, he has some problems that he labels “good problems”. I presume he believes these are good to get students thinking, so with that, let’s look at one:
Imagine a typical 6-sided die, and notice that the sum of opposite faces is always seven. The one is across from the six, the two is across from the five, and so on. Now imagine that you were making your own six-sided die that did not have this restriction. How many different dice could be made?
When I looked at the problem, I drew upon my prior knowledge and figured out that for the first of the three pairs of numbers there are 15 ways to do so, because one calculates how many combinations of two numbers can be made from six numbers. Knowing the formula for combinations of n things taken r at a time as n!/(n-r)!( r!), I plug in the appropriate numbers and obtain 6!/4!2!, or 15. \
That was the extent of my prior knowledge. More knowledge and assistance came from my friend Rob Craigen, who when he isn’t being a right-wing red-necked Canuck bastid, bides his time as a math professor at U. of Manitoba. He points out that having counted the number of ways we can pick the first pair, leaves four more numbers from which to make two more pairs. Again, this is found by the combination of four numbers taken two at a time: 4!/2!2!, which is six. Two numbers remain, and there is only one way to combine those. So we have 15 x 6 x 1 = 90 ways of picking three pairs of numbers.
Prof. Craigen labels these three pairs A, B and C and points out that because there are 3! = 6 ways to permute these labels, we have overcounted by a factor of 6. Dividing 90 by 6, we arrive at the answer: 15 ways to divide into three unordered pairs in which the pairs are not given in a particular order.
But we’re not done. For each group of three pairs of numbers, there are exactly two configurations, that can be placed on a die. For example, say we have the pairs 1,2; 3,4; 5,6. In one configuration, the 1,3 and 5 are placed clockwise around the corner where they meet, and in the other configuration they are placed counterclockwise around the corner. That means there are 15 x 2, or 30 distinct dice that can be made from the six numbers.
The is a rather complex problem as you can see. It is presented without any indication or recommendation of what grades would be appropriate to introduce it. Although this problem deals with permutations and combinations, which are covered in an Algebra 2 class, the concepts of the pairings is a bit more involved and might be more appropriate for a college course in combinatorics. But given the popularity of Problem-Based Learning and other artifacts of the above-mentioned reform tsunami, I am probably not too far off in envisioning teachers presenting the problem in middle school grades—or lower.
Without the necessary background, the chances are that students presented with this problem, after collaborating with their partners at the whiteboard or other vertical surface, will resort to manually counting: 1,2; 1,3; 1,4; … etc. Such brute force approaches work to obtain an answer—and one which is only partially correct as shown above. Also, such method does not connect to the larger concepts of permutation and combination and how they are used.
Instead, teachers who have been indoctrinated in eschewing traditional methods of math education, see such problem-solving exercises as fostering algebraic thinking. At the risk of stating the obvious, such efforts do no such thing. It is students muddling their way through—the quicker students showing off at the board with others completely lost, and no techniques learned or any connections made. This last is all the more ironic given that the criticism levied against traditionally taught math is that procedures are taught in isolation without connecting to other concepts or ideas.
And the irony of ironies is that students when confronted with other problems for which they have little or no prior knowledge, will “mimic” the same brute force methods—and at the end of all that work, they may or may not obtain the correct answer. Teachers, having been told this is all wonderful, will then applaud, point to, and evangelize what is now considered as “thinking”.
