# Teaching algebraic expressions: Counting smileys

January 29, 2011 1 Comment

I love this task for introducing the concept of variable and algebraic expressions through problem solving. It combines numerical, geometric, and algebraic thinking. The figure below shows the standard version of the task. Of course some easier versions would ask for the 5th figure, then perhaps 10th figure, then the 100th figure, and then finally for the nth figure. This actually depends on the mathematical maturity of the students.

An alternative version which I strongly encourage that teachers should try is to simply show first the diagrams only (see below) with the following instruction: Study the figures from left to right. How is it growing? Can you think of systematic ways of counting the number of smileys for a particular “Y” that belongs to the group? This way it will be the students who will think of which quantity (maybe the number of smileys in the trunk of the Y or the position of the figure) they could represent with *n*.The students are also given chance to study the figures, what is common among them, and how they are related to one another. These are important mathematical thinking experiences. They teach the students to be analytical and to be always on the lookout for patterns and relationships. These are important mathematical habits of mind.

Here are possible ways of counting the number of smileys: The n represents the figure number or the number of smiley at the trunk.

1. Comparing the smileys at the trunk and those at the branches.

In this solution, the smileys at the branches is one less than those at the trunk. But there are two branches so to count the number of smileys, add the smileys at the trunk which is n to those at the two branches, each with (n-1) smileys. Hence, the algebraic expression representing the number of smileys at the nth figure is **n+2(n-1)**.

2. Identifying the common feature of the Y’s.

The Y’s have a smiley at the center and has three branches with equal number of smileys. In Fig 1, there are no smiley. In Fig 2, there is one smiley at each branch. In fact in a particular figure, the number of smileys at the branches is (n-1), where n is the figure number. Hence the algebraic expression representing the number of smileys is **1+ 3(n-1)**.

3. Completing the Y’s.

This is one of my favorite strategy for counting and for solving problems about area. This kind of thinking of completing something into a figure that makes calculation easier and then removing what were added is applicable to many problems in mathematics. By adding one smiley at each of the branches, the number of smileys becomes equal to that at the trunk. If n represents the smileys at the trunk (it could also be the figure number) then the algebraic representation for counting the number of smilesy needed to build the Y figure with n smiley at each branches and trunk is **3n-2**, 2 being the number of smileys added.

4. Who says you’re stuck with Y”s?

This is why I love mathematics. It makes you think outside the box. The task is to count smileys. It didn’t say you can not change or transform the figure. So in this solution the smileys are arrange into an array. With a rectangular array (note that two smileys were added to make a rectangle), it would be easy to count the smileys. The base is kept at 3 smileys and the height corresponds to the figure number. Hence the algebraic expression is (3xn)-2 or 3n-2.

The solutions show different visualization of the diagram, different but equivalent algebraic expressions, and all yielding the same solution. Of course there are other solutions like making a table of values but if the objective is to give meaning to algebraic symbols, operations, and processes, it’s best to use the visuals.

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