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.

2 thoughts on “Revisiting the old AMC 10 Counting problem

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