# Emma's Dilemma

Extracts from this document...

Introduction

EMMA’S DILEMMA

I am going to investigate the number of different arrangements for the letters in the name LUCY

1 arrangement

LUCY

A different arrangement is:

LUYC

Another arrangement is:

LCUY

This combination consists of all the letters being different. I will also try and find a formula for this arrangement, and number of letters. Once I have discovered these formulae I am going to investigate, other combinations of letters and different amounts of letters. I will then try to discover a link between the formulae to enable me to find a formula for the general case.

ARRANGEMENTS IN THE NAME LUCY

LUCY UCYL CYUL YCUL

LUYC UCYL CYUL YCUL

LCUY ULCY CULY YUCL

LCYU ULYC CUYL YUCL

LYUC UYCL CLYU YLCU

LYCU UYLC CLUY YLUC

There are 24 different possibilities in this arrangement of 4 letters all different. I have noticed that with Lucy beginning each different letter. For example there are 6 arrangements with LUCY beginning with L, and 6 beginning with u and so on 6*4(the amount of letters) gives 24. This means that all names with 4 different letters will have a total of 24 different combinations.

I realised that the total number of combinations is the number of letters multiplied by the previous number of combinations.

Name: Ian

Total number of letters: 3

Middle

120

6

720

7

5040

From the table of results I have found out that a 2-letter word has 2 arrangements, and a 3-letter word has 6.

Taking for example a 3-letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4!, which is called 4 factorial which is the same as 4 x 3 x 2.

So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8-letter word.

###### Why this Formula Works

This formula works on the simple principle that once a letter is selected, the next selection will have 1 less choice, and so on.

For example: Lucy has 4 different letters.

Out of the 4 letters one letter has to be chosen to be the first letter, the second letter will be one of 3 possibilities because one has already been chosen (4 – 1 = 3). The third letter will be chosen out of 2 possibilities and the forth there will only be 1.

So then the possibilities have to multiplied together to achieve the total number of combinations.

4 x 3 x 2 x 1

So the original number of letters is the number of letters factorial.

Now I am going to investigate the number of different arrangements in a word with 2 letters repeated

EMMA

ARRANGEMENTS IN EMMA

EMMA EMAM EAMM MMAE

MMEA MEAM MEMA MAME

MAEM AMME AMEM AEMM

4-letter word, 2 letters repeated, 12 different arrangements.

Total number of arrangements is 12.

Having found a formula for Lucy when all the letters are different, I am now going to try to develop the formula for Emma, when 2 of the letters are the same.

If you look at four-letter word Lucy got 24 arrangements, but if you look at the name Emma it got four letters as well but it got 12 arrangements. This is because it got 2 letters the same.

I looked at other names with different total numbers and put the results in the table below.

## 2 Letters the same | ||

Name | Total number of Letters | Total Combinations |

Nn | 2 | 1 |

Ann | 3 | 3 |

Emma | 4 | 12 |

Jenny | 5 | 60 |

Conclusion

z = number of repeated letters (2).

If I go back to xxxyy; there are 3 x's and 2 y's in a total of 5 unknowns. As each letter has its own number of arrangements i.e. there were 5 beginning with x, and 5 beginning with y, I think that factorial has to be used again. Also in a 5-letter word there are 120 arrangements and 24 arrangements (120 divided by 5) for each letter. As there I a divide issue involved I had a go at trying to work out a logical universal formula. I came up with; The number of total letters factorial, divided by the number of x's, y's ect factorised and multiplied.

For examples:

A five letter word like aaabb; this has 3 a's and 2 b's. So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10

A four letter word like aabb; this has 2 a's and 2 b's

So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6

A five letter word like aaaab; this has 4 a's and 1 b

So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5

Five letter words like abcde; this has 1 of each letter (no letters the same)

So : 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24

All these have been proved in previous arrangements. This shows that my formula works.

Total letters | Number of A’s | Number of B’s | Number of arrangements |

3 | 2 | 1 | 3 |

4 | 2 | 2 | 6 |

5 | 2 | 3 | 10 |

6 | 2 | 4 | 15 |

3 | 3 | 0 | 1 |

4 | 3 | 1 | 4 |

5 | 3 | 2 | 10 |

6 | 3 | 3 | 20 |

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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