Wednesday, October 19, 2011

Now what? Part IV

So far in this series I have discussed the need to empower learners by getting them to ask and explore their own "Now what?" questions (here), considered possible answers to a messy learner-generated word problem (here), and identified implicit conditions associated with the different answers (here). In this final post of the series, I share my preservice teachers' efforts to extend our understanding of one of the possible answers to this word problem:
In the Community, you get two pets. The Elders pick the pets for each family. There were six choices of pets to have: dog, cat, fish, snake, bird, and hamster. What was the probability of getting a dog and a cat?
Three of the possible responses represent combinatoric approaches that the preservice teachers are already familiar with: combinations, permutations, and the multiplication principle. The 1/21 answer (duplicates are allowed but order does not matter), however, represents a new approach.

Many of the learners decide to explore, "How could I generalize this result?" They quickly realize that the 21 comes from adding the combination, 6 choose 2, with the 6 pairs. Consequently, they hypothesize that the general case would be:
n choose 2 + n
In order to check this rule, they try out some simpler problems. 1 pet results in 1 possible pair. 2 pets result in 3 possible pairs. 3 pets result in 6 possible pairs. 4 pets result in 10 possible pairs. 5 pets result in 15 possible pairs.

While this satisfies many of them, a few embrace the idea of extending the problem and notice that the sequence 1, 3, 6, 10, 15, 21, ... looks familiar. In fact, it can be thought of as:
(n + 1) choose 2
This fascinates them and me. I knew this result going into the lesson but I resisted the urge to explore it further before the lesson. "Why does this work?" was a question I did not have an answer to ahead of time. I wanted to work with them instead of guiding them to the answer. (This is an important instructional approach that I wrote about here.)

We are able to show that the two approaches are equivalent fairly easily.
But this still does not explain why (n + 1) choose 2 works when selecting 2 pets from n animals allowing for duplicates.

Finally we begin listing possibilities which leads to us designing the following table. The first entry would be a pair of dogs.
The "+ 1" is the repeat column in the table. This satisfies the "Why?" question but leads us to consider what would happen if 3 pets were selected from 6 animals - allowing for duplicates. Time is up, however, meaning this will be something we can think about on the drive home. Enjoy!


  1. If the pets were selected by rolling a die then the probability of getting a dog and a cat would be 2/36. I'm surprised that nobody picked that answer.

    There is a very nice formula for the number of ways to select k items from a group of n items with duplicates allowed. I won't spoil it here, but I will refer the reader to the Wikipedia article on the "Stars and bars" method.

  2. Nice catch, Dave. They actually did come up with 2/36 once they considered how the pets are being picked - another condition they had to account for. But this series of posts was getting long and I have a stack of other ideas I want to get to.

    Thanks for the suggestion about "Stars and bars." I'll be sure to check it out.