This problem was set to the Year 6-7 group in the Naturally Mathematical Challenge of 2003. While there were many excellent solutions to this question, none used the kind of systematic approach that we would advocate. It goes like this:

Step 1: Consider the positions of that the counter can take in the first row:

We have also labelled the positions that the counter in Row 2 can take, and these will have to be considered separately.

Step 2: Note that the last two positions in Row 1 are just the 'flip' of the first two positions, and so we can use the results of the first two positions for them.

Step 3: Systematically explore the positions for the second row. The results are:

In these diagrams, A, B and C show alternatives on the one diagram and the numbering refers to the position of the counter in Row 2.

Step 4: Summarise the results in a table, using the idea from Step 2 to make positions 4 and 5 the same as positions 2 and 1:

This gives us 24 different arrangements. The fact that there just happen to be 24 triangles in this T-Hexagon is, we think, nothing more than a coincidence!

Finally, what happens if we exclude arrangements that can be made from others by rotations or reflections?  When we did this, we discovered that only three essentially different solutions remain, and all these are shown with position 1 in row 1 filled:

Problem 1A

What is the largest number of objects that can be positioned in a T-Hexagon of side 2 such that there is one of each colour in every row?

To make this arrangement, we used 2 of the first and third configurations and one of the second. There may be other ways of meeting the constraint that there has to be one object of each colour in every row. However, note that the first configuration is the only one that uses a space in the inside hexagon. Since only two of this configuration can be positioned in a T-Hexagon of side 2, the above solution can't be bettered in terms of number of objects used.