In this project I hope to achieve the ability to explain how and why certain formulae can be associated with this particular problem. I also hope to extend the investigation in order to find interesting patterns, which can be relevant to the task.

The main variables that can be experimented with are: -

The number of colours used to shade the rectangle

The number of squares shaded in the rectangle

The number of squares in the rectangle

In order to investigate the problem thoroughly, it is sensible to only change one variable at a time whilst keeping the others constant.

First let’s investigate with the number of colours shaded, keeping the number of squares in the rectangle at 6. We will use a different colour each time so that the total number of colours used remains at 6 as well.

With 1 colour:

There are 6 places in which the red square can be placed.

Therefore there are 6 combinations

With 2 colours:

There are 6 places in which the red square can be placed. Once it is fixed and cannot move, there are 5 remaining places that the blue square can be placed.

Therefore there are 6 × 5 = 30 combinations

With 3 colours:

There are 6 places in which the red square can be placed and 5 remaining places that the yellow square can be placed. Once both colours are fixed, there are 4 remaining places that the yellow square can be placed.

Therefore there are 6 × 5 × 4 = 120 combinations

With 4 colours:

Once the red, blue and yellow squares have been fixed (120 possible combinations), there are 3 remaining places in which the green can be placed.

Therefore there are 6 × 5 × 4 × 3 = 360 combinations

With 5 colours:

Once the red, blue, yellow and green squares have been fixed (360 possible combinations), there are 2 remaining places in which the grey can be placed.

Therefore there are 6 × 5 × 4 × 3 × 2= 720 combinations

With 6 colours:

There is one remaining place in which the brown can be placed after the other 5 are fixed.

Therefore there are 6 × 5 × 4 × 3 × 2 × 1 = 720 combinations

However, the reason why the answer for 5 colours and 6 colours are the same is because the yellow remaining square is interchangeable with the 1 white square in the example above.

The results we have achieved from investigating the number of squares shaded using different colours are: -

6, 30, 120, 360, 720, 720

What type of results would we obtain if we used the same colour each time we added another square to be shaded? For a start we can see that the squares 5 + 1 and 1 + 5 count as the same combination. Let’s see what else we can investigate.

To show the different combinations, I have numbered each of the squares of the rectangle to make it clearer of the possible combinations.

With 1 black square shaded:

There are 6 possible places that the red square can be placed.

With 2 red squares shaded:

Possible combinations: 1, 2 2, 3 3, 4 4, 5 5, 6

1, 3 2, 4 3, 5 4, 6 5, 6

1, 4 2, 5 3, 6

1, 5 2, 6

1, 6

In total there are 15 possible combinations.

With three red squares shaded:

Possible combinations: 1, 2, 3 2, 3, 4 3, 4, 5 4, 5, 6

1, 2, 4 2, 3, 5 3, 4, 6

1, 2, 5 2, 3, 6 3, 5, 6

1, 2, 6 2, 4, 5

1, 3, 4 2, 4, 6

1, 3, 5 2, 5, 6

1, 3, 6

1, 4, 5

1, 4, 6

1, 5, 6

In total there are 20 possible combinations

With 4 red squares shaded:

Possible combinations: 1, 2, 3, 4 2, 3, 4, 5 3, 4, 5, 6

1, 2, 3, 5 2, 3, 4, 6

1, 2, 3, 6 2, 3, 5, 6

1, 2, 4, 5 2, 4, 5, 6

1, 2, 4, 6

1, 2, 5, 6

1, 3, 4, 5

1, 3, 4, 6

1, 3, 5, 6

1, 4, 5, 6

In total there are 15 possible combinations.

The reason why shading 4 red squares and shading 2 red squares is the same is because by shading 4 red squares, we leave 2 white squares remaining. Therefore by shading 2 red squares, we leave 4 white squares remaining. This is why both 2 ...