# Emma&#146;s Dilemma.

Extracts from this document...

Introduction

Abdul K Khan Mathematics Coursework 20/04/2007

Emma’s Dilemma

Introduction

In this mathematics coursework, I am investigate and find out the different combinations in a word after I have done this will be trying to find out a formula that will allow me to find out the different combinations quicker without showing all the different outcomes, and what I am also going to do is find a formula which will enable combinations easier for a word with repeated letters.

Different arrangements in the word LUCY (four letter word)

LUCY | ULCY | CLUY | YLUC |

LUYC | ULYC | CLYU | YLCU |

LCUY | UCLY | CULY | YULC |

LCYU | UCYL | CUYL | YUCL |

LYUC | UYLC | CYLU | YCLU |

LYCU | UYCL | CYUL | YCUL |

Number of combinations= 24

What I did to find out all the outcomes in the word LUCY was keeping the first two letters in the same place then just swapping the last two letters around until I was enable to find any more

Letter(s) | Number of letters | Total combinations |

L | 1 | 1 |

LU | 2 | 2 |

LUC | 3 | 6 |

LUCY | 4 | 24 |

1×2×3×4=24 combinations for the word LUCY

Different arrangements in the word EMMA (four letter word with same letters)

EMMA | MEMA | AMME |

EAMM | MAME | AEMM |

EMAM | MMEA | AMEM |

MMAE | ||

MEAM | ||

MAEM |

Middle

0

1

SS

2

2

2

1

1 × 2 ÷ 1 ÷ 2 = 1 combination for the word SS

Formula meanings

n = number of letters in the word

y = number of repeated letters

! = factorial

My formula

My formula for word with different letters is

n! =

n × (n – 1) × (n – 2) . . . 2 × 1

This is because it is quicker to find the total combinations for a certain amount of letters without writing them all up.

The formula for letters that have been repeated in the word is

n! ÷ y!=

n × (n – 1) × (n – 2) . . . 2 × 1

y × (y – 1) × (y – 2) . . . 2 × 1

This is because without dividing you will get some same combinations when written up.

Total combinations for words with letters up to ten

Number of letters | formula | Total combinations |

1 | 1! = | 1 |

2 | 2! = | 2 |

3 | 3! = | 6 |

4 | 4! = | 24 |

5 | 5! = | 120 |

6 | 6! = | 720 |

7 | 7! = | 5040 |

8 | 8! = | 40320 |

9 | 9! = | 362880 |

10 | 10! = | 3628800 |

Total combinations for words with repeated letters up to ten

Z= total number of combinations with same letters

Number of letters | What you can get when different letters | Number of repeated letters | Total combinations |

1 | 1 | n! | z |

2 | 2 | n! | z |

3 | 6 | n! | z |

4 | 24 | n! | z |

5 | 120 | n! | z |

6 | 720 | n! | z |

7 | 5040 | n! | z |

8 | 40320 | n! | z |

9 | 362880 | n! | z |

10 | 3628800 | n! | z |

Extension

What I am going to do now is find a formula for a word with different sets of same letters.

Different arrangements for the DEDE (four letter word with sets of same letters)

DEDE | EDED |

DEED | EDDE |

DDEE | EEDD |

Number of combinations= 6

Letter(s) | Number of letters | What you can get when different letters | Number of repeated letters that are same | Number of repeated letters that are same | Total combinations |

D | 1 | 1 | 0 | 0 | 1 |

DE | 2 | 2 | 0 | 0 | 2 |

DED | 3 | 6 | 2 | 0 | 3 |

DEDE | 4 | 24 | 2 | 2 | 6 |

Conclusion

This is because it will make it easier with things with more letters and you may also write down some same combinations.

This is a table showing more set of letters in a certain amount of letters up to ten

Number of letters | What you can get when different letters | sets of repeated letters | Total combinations |

1 | 1 | n! ÷ n! . . . n! ÷ n! | z |

2 | 2 | n! ÷ n! . . . n! ÷ n! | z |

3 | 6 | n! ÷ n! . . . n! ÷ n! | z |

4 | 24 | n! ÷ n! . . . n! ÷ n! | z |

5 | 120 | n! ÷ n! . . . n! ÷ n! | z |

6 | 720 | n! ÷ n! . . . n! ÷ n! | z |

7 | 5040 | n! ÷ n! . . . n! ÷ n! | z |

8 | 40320 | n! ÷ n! . . . n! ÷ n! | z |

9 | 362880 | n! ÷ n! . . . n! ÷ n! | z |

10 | 3628800 | n! ÷ n! . . . n! ÷ n! | z |

Conclusion

In this coursework I have discovered that the letters in a word or a lot of letters put together can be re-changed to find different combinations. For with and without same letters I have a formula to make it easier to the total of combinations without the working out. I have also discovered it would work other situation e.g. when there is more sets of same letters.

- -

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

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month