Discrete Math - Day 14

Today we continued to work on the grouping problem from last class. Students did not have any new insights. I figured something like this would happen and took this opportunity to re-iterate that looking at smaller problems helps to get a sense of the problem dynamics.

I had three people stand at the front of the room. I asked the class how many ways could I break these three people into three different groups? After gaining clarification that reordering doesn't matter, students determined that there was only one way to form three groups.

I had another student join the other three up front. We now had four students and we wanted to form one group of 2 and two groups of 1. The question was how many ways could this be done? To emphasize what was going on, I had the students rearrange and pair themselves in different ways.

Students started to perform mental calculations and were thinking it was 6 different ways. Then some students questioned that result because the felt the number of people in the two groups of 1 were not being accounted for. There were a lot of questions and uncertainty coming out of the discussion.

I stepped in and asked students how many people were we choosing from to form the first group? Students responded that there were four. I then asked how many of these four were we choosing to form the first group. The response was 2. At this point students started to ask whether this meant that they could use ₄C₂. I asked students to calculate this value and they found it matched their count of 6.

I then asked what would happen with the two groups of 1? There again was some confusion but students finally resolved that those two groups were now fixed since there were only two people left to fill the two empty slots.

We then moved to five people and first looked at one group of 2, one group of 2, and one group of 1. Students started to pick up that there would be ₅C₂, ₃C₂, and 1 ways of selecting each group. There were then discussions about whether to multiply or add these values.

I related this to having 4 shirts/blouses and 3 pairs of pants. Students knew that they would have 12 outfits. When asked why they responded that was what they were told to do. At last someone said that every time they selected a pair of pants they would have four options for their top. This made sense to the class.

I then asked the class what that meant for the group situation. There was still some confusion. I asked how this problem was related to the pant-top situation. For the first group of 2, this group could be matched to other groups formed. I again used the students at the front of class to illustrate what was going on.

I asked students to sit down and now looked at forming groups with 6 people. The first option was one group of 3, one group of 2, and one group of 1. Students started to see that there would be ₆C₃, ₃C₂, and 1 option.

We then tackled three groups of 2. Here students thought there would be ₆C₂, ₄C₂, and 1. Someone asked if the last group should be calculated as ₂C₂, which is technically what is happening. I agreed that that best represents the situation. When students calculated the result was still one.

In the course of this discussion students noticed that ₅C₃ and ₅C₂ came out to the same value and wondered why. We took a more in-depth look at how permutations and combinations are calculated. I also related back to the connection between combinations and Pascal's triangle with the idea that the symmetry in the triangle will manifest itself as symmetry in values. Students could see how the formula for calculating combinations would lead to this symmetry.

For those of you who are not familiar with these formulas, we have

After going through these explorations, I had students again turn to the original problem from last class. Students found 792 ways to form groups of 7 from 12 people. They also found they could form groups of 3 from 5 people 10 different ways and concluded that the last group of 2 could only be formed in one way.

When I asked people what the final number was, students either calculated 803 or 7920. This was a situation where students had added, so there was still some confusion on this point. We discussed this again and students decided that 7920 was the correct result.

Some students said they started using the groups of 2 and 3 first, i.e. they calculated ₁₂C₂ and ₁₀C₃. They were concerned that they would get a different answer. We examined this and students started to see that the results all turned out the same and that the symmetries present in the calculations were coming into play.

I then had students work on a second grouping problem. For this problem everyone seemed to readily grasp what to do and the problem was quickly dispatched.

With a few minutes remaining, I previewed the ideas of counting and probability. Discrete probability will be the focus of the next couple of class.

Visit the class summary to read a student's perspective of the class and to view the lesson slides.