# Emma's Dilemma

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Introduction

GCSE Maths Coursework

Emma’s Dilemma

Introduction

Emma and Lucy are playing with arrangements of the letters of their names, on arrangement of Lucy is:

LUCY

Another arrangement is:

YLCU

I will be investigating all the various arrangements of ‘Lucy’ and all the arrangements possible for ‘Emma’

I will also investigate the various arrangements of words with different lengths and different letters, i.e. 3 lettered words, 4 lettered words, 5 lettered words, etc. I hope to then find a pattern which will then help me take my investigation further. I will split my investigation into three parts to make it more systematic and easier to understand. The first part will consist of me finding the various arrangements of ‘Lucy’ by listing them all. The second part will consist of me finding the various arrangements of ‘Emma’ which I will also do by listing. Finally the third part will consist of me trying to find out how many possible arrangements there are for words with various lengths and different letters, as I mentioned above (3 lettered words…).

Part 1

I will be listing all the various possibilities for ‘Lucy’

Lucy

Whilst listing the different possibilities I will try to be as systematic as I possibly can.

LUCY ULCY CLUY YLCU

LUYC ULYC CLYU YLUC

LCYU UCLY CYLU YULC 24 POSSIBILITIES

LCUY UCYL CYUL YUCL

LYCU UYLC CUYL YCUL

LYUCUYCL CULY YCLU

In the effort of trying to be systematic, whilst I was listing the various arrangements I decided to split it into four parts, each part for a different letter of the four lettered name. This is indicated above.

Part 2

I will now be listing all the different possibilities for ‘Emma’

Emma

Whilst listing the various possibilities I will once again try to be as systematic as I possibly can.

Middle

2 * 3

= 6 NUMBER OF POSSIBILITIES

I have colour coordinated this calculation so that I can explain it. The 2 is the previous answer I obtained, which can be checked in the table plotted above. The 3 is the number of letters in the word, and the 6 indicates all the various possibilities. As you can see, this answer is correct and can be double-checked in the table above.

I will also do another example to double-check that my method works, and that the pattern is correct. Therefore I will be using the four lettered word, and the outcome should be 24, as I have worked this out previously using the listing method. My working is shown below:

6 * 4

= 24 NUMBER OF POSSIBILITIES

This proves that my pattern is correct, and that the method works. This means that I can add on to my table of discoveries, using my method of working. Using the method I will now do a five lettered word, and then a six lettered word. My working will be shown below:

Five Lettered Word

24 * 5

= 120 NUMBER OF POSSIBILITIES

Six Lettered Word

120 * 6

= 720NUMBER OF POSSIBILITIES

I will now show you a development of my table.

Number Of Letters In The Word | Number Of Possible Outcomes |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

6 | 720 |

Whilst doing my working I also managed to find another pattern. This is shown on the following page.

1 2 3 4 5 6

1 2 6 24 120 720

1 * 1 * 2

1 * 1

1 * 1 * 2 *3

1 * 1 * 2 * 3 * 4

1 * 1 * 2 * 3 * 4 * 5

1 * 1 * 2 * 3 * 4 * 5 * 6

I then showed all my discoveries to my teacher, who then pointed out that what I had worked out is known as a factorial. I was then introduced to the factorial button on the calculator. This would enable me to workout the number of arrangements possible for a word with N number of different letters. I have come up with a conclusion of:

N!

Conclusion

4! / 4!

Therefore I decided to multiply them together, hence giving me the equation mentioned above:

4! / (2! * 2!)

= 6

The only way to prove whether or not this is correct is by listing all the possible outcomes, and hence I have done so on the following page.

AABB

AABB BBAA

ABAB BABA 6 POSSIBILITIES

ABBABAAB

I tried to be as systematic as I possibly could, and the number of outcomes is indicated above.

According to this my prediction seems to be correct, and hence I have managed to come up with another generalised equation, which can be used to work out the various combinations possible with a word with any length and more than one letter repeated. This is shown below:

AAA…ABBB…B

ST

N!

(S! * T!)

An example is ‘AAAABBB’

7! / 4! * 3!

= 35NUMBER OF POSSIBILITIES

The equation can also be extended if necessary. This is done if there are more than two letters that are repeated, for example ‘AABBCCDD.’ The equation in this case will be:

N!

OR 8! / (2! * 2! * 2! * “

(S! * T! * U! * V!) =2520

Conclusion

In conclusion I managed to find three different formulas which are developments of each other. As I worked on each part of the investigation I added a little part to the equation, and by the end of part five I have ended up with the final equation. This can be used to find all the possible combinations of any particular word, no matter how many repeats it has, nor how many letters are repeated. For example I can work out all the possible outcomes for a random word such as ‘MISSISSIPPI’ a shown below. Before I work out the suggested word I will show you the evolution of my equation.

- N!
- N! / S!
- N! / (S! * T!)

The working used to find how the number of possible outcomes for the word ‘MISSISSIPPI.’

Equation = N! / (S! * T! * U!)

= 11! / (4! * 4! * 2!)

= 34650

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

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