Wednesday, 20 September 2017

8 in a row

There is a book called "Open Ended Mathematics Activities" by the godfather of Australian mathematics education, Professor Peter Sullivan. If you do not have a copy, you need to get one. (It is available through AAMT - try here: 

There is an activity in the book that asks you to imagine a line of 8 students, some of them standing up and some of them sitting down. How many different ways could you organise the standing and sitting children?

We started working on this problem. The kids got several different solutions so I stopped them and asked them to predict how many different ways they thought it could be done. Most answers ranged from 4-8. One student predicted 10 solutions, another 20 and a final student guessed 1000. 

"He probably just means a really big number," said one perceptive student.

I decided to change the task a little bit and break it down into a few smaller steps.

So I took it back to the simplest possible question:

If there are 8 children and only 1 of them is sitting down, what are all the possible positions that the children could be organised in?

The first couple of minutes we were pretty random in our strategies. We were just playing with combinations and seeing what happened.

Then the "A-ha!" moment when one group discovered the advantages of being organised and systematic.

We found that there were 8 possible ways for the line of 8 to be organised with one sitting and 7 standing.

So I asked, what if there were 2 sitting - but they have to be sitting together? (I asked for them to be sitting together to try to simplify the problem and to limit the possibilities. It also revealed some interesting patterns - see below.)

Once we had seen how to "get organised" and systematic, it was a pretty quick journey to finding that there were 7 possible arrangements with 2 sitting together and 6 standing.

Here's one for those Richmond fans...

Three sitting and 5 standing.

Then with a bit of intuitive thinking, some of the students saw a pattern emerging.

The number of possible solutions plus the number of sitting students will always equal 9 (one more than the number of students).

For example:

And yes it proved to be true.

We only needed to go a few more steps before we were convinced.

Yep - 4 sitting plus 5 combinations equals 9.

So adding up all the possible combinations, we found 36 ways to arrange 8 students, some standing and some sitting. But these were only the combinations where the sitters are sitting together.

Now, I wonder how many possible combinations there would be if the sitters could be ANYWHERE in the line?

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