Discrete Math - Day 13

Well, after yesterday's outstanding class, things were a little flat today. It sometimes takes a little more to get things going in a first period on a Friday morning.

We started out looking at how many ways five cards could be drawn from a deck. We compared this result with the possibilities when drawing four cards from yesterday. Students are amazed that there are over 2.5 million different hands that can be drawn. We used the full-house count we derived to show that probability of being dealt a full-house is very, very small. Although we haven't talked about discrete probability formally yet, this is our next topic, so I like to start introducing the connections between the counts we are performing and calculating probabilities.

We then looked at a grouping problem. This is where things really started to stall out. Students tackled the problem, but without much energy. The problem asks to split 12 people into one group of 2, one group of 3, and one group of 7. Many approached the idea that combinations were involved. The issue came in trying to figure out what to do with these values.

I used a messenger service and had one member of each group rotate through several other groups so they could get different perspectives on whether you should add or multiply values.

Students were still stuck but one student did look at a smaller problem. He looked at splitting 5 students into one group of 2 and one group of 3. He listed out the group of 2 options and saw that it was 10, which fit his use of ₅C₂. He also saw that there were 10 options for forming the group of 3. He couldn't figure out why they must be equal.

I had him share his work with the class and asked them why if there were 10 groups of 2 that we must have 10 groups of 3. Some students thought that it was because the 5 was an odd number. They had looked at ₆C₂ and ₆C₃ and these were different. I then posted up several combinations that I told them would be equal and asked why they must be equal.

     ₆C₄ = ₆C_{2     15}C₃ = ₁₅C_{12     7}C₁ = ₇C₆

I again asked students why these must be equal. Students observed that the groups being formed summed to the total but that's as far as they got.

At this point the class was ending. I asked students to think about what was happening in the formation of the groups, to connect the idea to the physical reality of what was happening as groups were formed.

Their homework was to try to figure out why these values must be equal and how this idea connected to the grouping problem that we were working on.

Hopefully some students will actually spend some time on the problem and come back with some ideas. This discussion will continue on Monday.

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