Discrete Math - Day 6

Today's focus was on building conceptual understanding of permutations. Before diving into this I wanted to cover an open question from the last lesson; "For linear equations, why does taking the difference of successive y-values result in the coefficient of the x-term?" I briefly explained using the values of n and n + 1 along with the equation y = 3x + 7 to show what happens when the two y-values are subtracted. I did this more to show how generalized values can be used to explain results. This is starting to build exposure to proofs that will come later in the course.

The two permutation problems we tackled today focus on permutations. Some students may have exposure to these ideas but few if any have a real comfort level with the ideas.

I start the class picture problem by having 4 students stand in front of class. I proceed to rearrange them several times so the class has a clearer idea of what is happening. Students worked on the problem and several groups thought the answer would result in N x N. Since we were dealing with N=30 it was hard to verify if this was correct. I referenced the four students at the beginning and asked if they could verify their conjecture using 4 students. I tried to emphasize with the class that you need a way to check if your results are correct and working with smaller values is one way to do this.

A few students made lists for smaller numbers and arrived at the conclusion that factorials were being used. These visuals helped make it clearer for everyone in the class why factorial accurately represented the situation.

Part A of the next problem is a good check to see if students understand what is going on. Students quickly identified that there would be 10! arrangements. Part B of the second problem throws in a wrinkle by not using all of the items. You now are faced with having 5 items but only using 3 in the arrangement. This proved challenging for students. Several students came up with arguments as to why it could not be 5! or 3!, in effect creating bounds for the solution. A couple of students developed nice visual representations that made it clear the answer was 60.

One student represented the problem by making a list of all the arrangements for a fixed starting value (there were 12) and since any of the 5 items could be in the first position this would result in 60 total arrangements. A second student made a tree diagram and show there were five spokes to start, each of these had 4 spokes, and each of these had 3 spokes, resulting in 60 arrangements.

I had students conclude by writing down their thoughts about what to keep in mind when working with permutations and ways they can represent the problems.

Next class will dive into problems involving combinations.

Visit the class summary for a student's perspective and lesson slides.