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• Level: GCSE
• Subject: Maths
• Word count: 2718

# 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.

Extracts from this document...

Introduction

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.

Middle

Once again on first glance of these values, it does not appear to have a set pattern. However by thinking carefully how the values were achieved in the first place, it is possible to find one.

For example:        When we shaded 1 square there were 6 combinations

When we shaded 2 squares there were 6 × 5 combinations

When we shaded 3 squares there were 6 × 5 × 4 combinations

When we shaded 4 squares there were 6 × 5 × 4 × 3 combinations

From this pattern, we can correctly deduce that the formula will included a factorial sign. This means that a value such as 6 factorial will be multiplied by every whole number integer below it.

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Because the value for every number of shaded squares begins with 6 x, the formula must contain 6!.

For the remaining squares

1     =     1!

2     =     2!

6     =     3!

24   =     4!

120 =     5!

720 =     6!

Therefore the formula to find the number of combinations using separate colours is: -

However, r = n – s. Therefore the overall formula is: -

### To find the number of combinations using the same colour: -

6        15        20        15        6        1

The formula for the number of combinations using different colours is

This gives us the values: 6, 30 120, 360, 720, 720. To find how the two sets of data are related, we can divide one by the other to get a new row of values.

gives us: -

s                        1                 2            3             4              5            6

c (sep. colours)       6           30          120         360         720         720

c (same colours)     6           15           20           15            6             1

÷ (d)                        1            2             6            24           120         720

If we look at s and d

Conclusion

However, if another colour is added into the grid, the number of ways of arranging in a line n objects of which p of one type are alike, q of another type are alike, r of another type are alike and so on, is

Here are some examples of where this formula may come in use: -

Problem 4:        Using the word MATHEMATICS, how many ways can the letters be arranged?

Solution:        In the word MATHEMATICS, there are: -

11 letters and M occurs twice

A occurs twice

T occurs twice

Therefore, the number of ways that the letters can be arranged

=

=

= There are 4989600 ways of arranging the letters from the word  MATHEMATICS.

Problem 5:        Still using the word MATHEMATICS, what is the probability that having rearranged the letters once, the combination begins with MM?

Solution:        The probability space         S = (arrangements of MATHEMATICS)

N(S) = 4989600

Let Y be the event “the arrangement begins with MM”.

If this happens, we must be left with MMxxxxxxxxx. There is only one way that MM can be arranged. Therefore the remaining 9 letters can be arranged in  = 90720 ways.

Therefore n(Y) = 90720

P(Y) =

=

Therefore the probability that the arrangement begins with MM is

In conclusion to my investigation, I have explained why and how many formulae can be related to this particular problem. I managed to find patterns, test them and explain why they worked, which is what I set out to do.

Matthew Arnold

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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