Counting, Pascal’s triangle, and binary numbers

In yesterday’s counting project the boys noticed a connection between a counting problem and binary numbers. Here’s that project:

Revisiting an AMC 10 Counting Problem

For today’s project I wanted to explore that connection an a little more depth. To start off we looked at the connection between counting arrangements and Pascal’s triangle:

In the first part of this project we saw a connection between counting pairings of tourists and guides has a interesting connection with Pascal’s triangle. Here we look more carefully at that connection by trying to understand how the rule that tells you how to construct the rows of Pascal’s triangle also shows up when you count these pairings.

We explored the connection here in two parts. In this first part we show the 6 ways that you can pair 4 tourists with 2 guides when each guide has 2 tourists. We also show the 3 ways to pair 3 people with 2 guides where the first guide gets 1 person and the 3 ways to pair 3 people with 2 guides where the first guide gets 2 people.

Now we are ready to find the connection between the two lists me made in the prior video. That connection is important because it shows that the same addition rule that gives the rows of Pascal’s triangle also applies to counting arrangements of certain sets, and therefore helps you understand why Pascal’s triangle helps you count those arrangements.

In the last part of the project we explore the connection between binary numbers and Pascal’s triangle. We do this using an example of 5 digit binary numbers (from 00000 to 11111). This connection allows you to see that the rows of Pascal’s triangle always add up to be a power of 2.

So, a nice little project showing some fun connections between Pascal’s triangle, counting, and binary numbers. Some of these connections are pretty deep and I certainly don’t expect that the boys will have understood every detail from this project. They did seem to have fun with it, though, and their understanding seems to have come a long way from when we worked through the AMC 10 problem earlier this week.

Revisiting the old AMC 10 Counting problem

Yesterday we struggled through an old AMC 10 counting problem:

A bit of a struggle with our counting project today

Here’s the actual problem:

Problem 12 from the 2007 AMC 10a

Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?

After talking a little bit about permutations and combinations this morning, I thought we’d give the problem a second try. Not surprisingly it was a little easier the second time through, though there were still some challenges. We ended by talking about how this particular problem is related to counting in binary.

First, though, a quick review of permutations and combinations:

After that quick review we jump into the AMC 10 problem. Our first approach involves counting the ways that 1 person can go with the first guide, then 2 people with the first guide, and so on. My younger son got a little confused about the case with 2 people, so we looked at the case extra carefully. He was confused about whether or not order mattered.

With the confusion about order (hopefully) sorted out, we continued counting through the remaining cases. Here we also have a nice chance to talk about some of the symmetry in this problem.

Next we looked at the problem from a different perspective – the perspective of the tourists. This approach allows us to get to the answer to this problem in a way that seems completely different that the first approach. One fun thing my older son noticed here is that it seems as though our approach is related to counting in binary.

Finally, we looked more carefully at the connection to counting in binary. This connection probably deserves a project all to itself, but even going through things quickly here was fun.

It felt as though this project went much better than yesterday’s project. People naturally snicker a little bit when you say it, but learning to count is really hard. Yesterday the confusion involved adding and multiplying. Today the confusion involved whether or not order mattered. Working through those two ideas isn’t necessarily super easy, but I think we made some good progress. Fun little project.