My older son and I were looking at this problem the other day:
Problem 12 from the 2007 AMC 10a
Here’s the problem:
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?
Since we are studying counting, I thought reviewing this problem would make for a good project with the boys this morning. Unfortunately the project didn’t go as well as I was hoping, but we did get to have a good conversation about the times when you need to add and the times that you need to multiply when you count.
We started the project by having my older son explain his original solution. His approach is fairly straightforward and gave a couple of different opportunities for my younger son to help out. I’m glad that my younger son was able to participate because there are some fairly challenging counting ideas here:
We ended the last video with my younger son being a little confused about whether or not to add or multiply when we’ve looked at some cases. I decided to take a look at a slightly easier problem so that we could actually list out all of the cases. I hoped this exercise would help him see that we needed to add in the last video. Unfortunately writing out the cases was a little harder than I thought it was going to be.
However, there’s some pretty good ideas from both kids right at the end about the relationship between the numbers we have on the board and Pascal’s triangle.
For the last part of the project I showed them a different way to approach the problem – looking at it from the perspective of the tourists. We had a little more confusion about adding and multiplying. I tried to clarify that we needed to multiply here, but I’m not sure my explanation helped.
So, a fun problem, but not such a great project. Can’t win them all 😦
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