# Emma's Dilemma

Extracts from this document...

Introduction

## Emma’s Dilemma

For our maths coursework we had to investigate the number of different word arrangements possible for names. The names (or letter arrangements) I started off with had letters that were only used once n each name and were not repeated, like “Lucy”. As I continued experimenting with the different arrangements, I began to use words with letters repeating, such as “Emma”. My goal was to find formulas that worked for the different arrangements.

I started off by working out the number of different arrangements in the name “Emma”. I shall now display the different arrangements of “Emma” that I found:

EMMA

EMAM

EAMM

MMEA

MMAE

MAME

MAEM

MEAM

## MEMA

AMME

AEMM

## AMEM

Here are the different letter arrangements for “Lucy”:

LUCY ULCY CLUY YLUC

LUYC ULYC CLYU YLCU

LCUY UCLY CULY YULC

LCYU UCYL CUYL YUCL

LYUC UYLC CYLU YCLU

## LYCU UYCL CYUL YCUL

Middle

NMTYI NITYM NTIYM NYITM

NMYIT NIYMT NTYMN NYTMN

NMYTI NIYTM NTYNM NYTNM

TMINY TIMNY TNMIY TYMIN

TMIYN TIMYN TNMYI TYMNI

TMNIY TINMY TNIMY TYIMN

TMNYI TINYM TNIYM TYINM

TMYIN TIYMN TNYMI TYNMI

TMYNI TIYNM TNYIM TYNIM

YMINT YIMNT YNMIT YTMIN

YMITN YIMTN YNMTI YTMNI

YMNIT YINMT YNIMT YTIMN

YMNTI YINTM YNITM YTINM

YMTIN YITMN YNTMI YTNMI

YMTNI YITNM YNTIM YTNIM

I displayed each number of letters in the way of a tree diagram, each time changing the last letter of the word, then working my way back along the word, until I eventually had the letters in reverse order. I then recorded the results in this table:

NUMBER OF LETTERS | DIFFERENT ARRANGEMENTS |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

6 | 720 |

7 | 5040 |

8 | 40320 |

9 | 362880 |

10 | 3628800 |

Here is how I found the rule of working out the number of arrangements for a word with no repeated letters. If you multiply the number

Conclusion

I used the same method that I used before and discovered that words with one set of repeated letters were exactly the same as the previous results, except that the number of arrangements was halved when the letters were repeated. Therefore the algebraic formula would be n!/2. I then had to find a rule that could be applied to words with no repeated letters, two repeated letters and then three repeated letters. According to my previous formulas, an arrangement of words with three repeated letters should have an algebraic formula of n!/3. I decided to display my results in the following table:

NUMBER OF LETTERS | NO REPEATS | TWO REPEATED LETTERS | THREE REPEATED LETTERS |

1 | 1 | - | - |

2 | 2 | 1 | - |

3 | 6 | 3 | 1 |

4 | 24 | 12 | 4 |

5 | 120 | 60 | 20 |

6 | 720 | 360 | 120 |

7 | 5040 | 2520 | 840 |

8 | 40320 | 20160 | 6720 |

9 | 362880 | 181440 | 60480 |

10 | 3268800 | 1814400 | 604800 |

Whilst looking for different rules I found that you can also get the same answer by using the algebraic expression n!/r!, which means that the number of letters factorized divided by the number of repeated letters.

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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