# Emma's Dilemma.

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Introduction

## Maths Coursework-

Emma’s Dilemma

For this piece of coursework I am going to investigate the number of different ways I can write a word, re-arranging the letters without having any repeats of the sequence.

After I have finished my investigations I will try and use my findings to draw together a formula which I could then use to find out how many ways a word can be written for any chosen word.

My initial step is to write the name ‘EMMA’ with as many different arrangements I can find.

## Part 1

## EMMA
| ## 7) MAME |

## EMAM
| ## 8) MEAM |

## 3) EAMM | ## 9) MAEM |

## 4) MMEA | ## 10) AEMM |

## 5) MMAE | ## 11) AMEM |

## 6) MEMA | ## 12) AMME |

Next I am again going to try a 4 letter word, but this time without repeats (no 2 letters the same) in it.

I predict that a 4 letter without repeats will have a lot more letter arrangements than the name EMMA which has ‘M’ repeated.

Part 2- I have chosen the name ANDY.

## 1) ANDY | ## 9) NYAD | ## 17) DYNA |

## 2) ANYD | ## 10) NYDA | ## 18) DYAN |

## 3) ADYN | ## 11) NDYA | ## 19) YADN |

## 4) ADNY | ## 12) NDAY | ## 20) YAND |

## 5) AYDN | ## 13) DNYA | ## 21) YNDA |

## 6) AYND | ## 14) DNAY | ## 22) YNAD |

## 7) NAYD | ## 15) DANY | ## 23) YDAN |

## 8) NADY | ## 16) DAYN | ## 24) YDNA |

## This is double the number of ways EMMA can be written.

## My first thought to get the number of arrangements is possibly by dividing a value (which I do not know at present) by the number of times a letter is repeated in a word. E.g. SARA would be Χ/2, where Χ is the unknown value and the 2 is because letter A is repeated twice.

Now, I am going to try a 4 letter word, XXXY, which has X repeated 3 times. I predict there will be less re-arrangements.

Part 3

## 1) XXXY | ## 3) XYXX |

## 2) XXYX | ## 4) YXXX |

Middle

## 12) MNDAY

## 20) MYDNA

## 5) MADYN

## 13) MDNAY

## 21) MYNAD

## 6) MADNY

## 14) MDNYA

## 22) MYNDA

## 7) MNADY

## 15) MDYNA

## 23) MYAND

## 8) MNAYD

## 16) MDYAN

## 24) MYADN

1) BOB |

2) BBO |

3) OBB |

## I will now try a 3 letter word, which has no repeats. After doing this I will try another 3 letter word, which has a letter, repeated twice. I predict there will be twice as many combinations for the word without repeats.

Part 5- ROB Part 6- BOB

## 1) ROB |

## 2) RBO |

## 3) BRO |

## 4) BOR |

## 5) OBR |

## 6) ORB |

## What I predicted was correct, I got the idea from EMMA, which has a letter repeated, having half the number of combinations than the name ANDY, which has no repeats.

I think there must be a link between how many repeats there are in the word and the number of arrangements.

## I will now try and predict

Conclusion

I shall now try a new formula which will account for words that have more than one letter repeated. the new formula is N!

n1!n2!n 3!………………

Where n1! is the number of times the first letter repeated and n2! is the number of times the second letter is repeated and whenn3! Is the number of times the third letter is repeated. The ……………… means more nx! which can be added depending on how many letters are repeated.

I will try the new formula with XXXYY.

= 5!

3! x 2!

= 10✓

Conclusion

## My new formula, works for any given word. The N! stands for the factorial of number of letters in the word. n1! is the number of times the first letter repeated and n2! is the number of times the second letter is repeated and so on. The ………… represents more nx! which can be added depending on how many letters are repeated.

I could use my formula to work out any word and my last example is going to be FINISHED.

##

= 8!

2!

= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

2 x 1

= 20160 arrangements

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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