# Emma's Dilemma

Extracts from this document...

Introduction

Emmas Dilema _ Jonathan Busby

This piece of coursewrk is to investigate the possible arrangements of Lucy and Emma. Also I have to investigate the number of different arrangements of various groups of letters. I will start my experiment off by investigating the two-lettered word JO.

## Two-Lettered Word

JO

OJ

JO has a total of two letters making up its word. Therefore its total amount of arrangements is that of tom.

Three-Lettered Word

TOM, TMO, MOT, MTO, OMT, OTM

I have a three lettered word (tom) which has a total of 6 different arangements, as i have shown above.

## Four-Lettered Word

LUCY UCLY CYLU YCLU

LUYC UCYL CYUL YCUL

LYUC UYLC CULY YUCL

LYCU UYCL CUYL YULC

LCYU ULYC CLYU YLUC

LCUY ULCY CLUY YLCU

By arranging the letters in this way, I found that the four-lettered word LUCY had a total of 24 arrangements. By showing how many arrangements there were when the single letter L was at

Middle

2 2

3 6

4 24

5 120

6 720

The number of different letters multiplied by the previous number of arrangements equals the total amount of arrangements for the next of letters. This can be shown as 720 x 6 .

This is a table to show this:

No. Of Letters (N) | No. Of Arrangements |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

6 | 720 |

7 | 2520 |

## Four-Lettered Word- Two of any one letter

EMMA MEMA MEAM AMME

EMMA MEMA MEAM AMME

EAMM MMEA MMAE AMEM

EAMM MMEA MMAE AMEM

EMAM MAME MAEM AEMM

EMAM MAME MAEM AEMM

I have found out that from any four-lettered word i will get 24 different arrangements.

I will check doubled words.

For a 4 Lettered word there are 24 arrangements in total, as my results show, next i will investigate is the results of having two of the same letters in any-one word and its effects on the outcome of the number of arrangements. I will firstly look at the four-lettered name Emma. As you can see two of its letters are the same therefore we have to treat them as one.

Conclusion

2520

7

5

7! ÷ 5!

7x6x5x4x3x2x1÷5x4x3x2x1

42

10

7

10! ÷ 7!

10x9x8x7x6x5x4x3x2x1÷7x6x5x4x3x2x1

720

This formula can be used to work out as many letter repeats, as you like.EXAMPLE

Banana

In the word banana there are 6 letters altogether making it 6x5x4x3x2x1, which is equal to the formula of 6! There are 3 A’s repeated and also 2 N’s. Also there is one b.! In the formula.

This is 6! ÷ 3! ÷ 2!

Conclusion/evaluation

It is possible to see from my research that overall as the number of letters in a word increases, the number of combinations increases. If a letter is repeated in a word, that word has less combinations than another word of equal length with no repeated letters. If i did the experiment again i would spend more time with more variety of names. i would use more double letters and look for anu other formulaes.This would mean that into more detail and check the formula for the larger words.

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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