# Emma's Dilemma.

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Introduction

Emma’s Dilemma

Introduction

This coursework is about rearrangements of the letters in peoples names. We will be looking at how many combinations you can make out of the letters in different size names. I will then work out a formula that will let me work out the number of combinations in a name given the number of letters in that name. I will work this out in a systematic way, by changing one letter at a time and moving down the word.

The names I will be using are:

2 letter name

BO

3 letter name

SAM

4 letter name

MIKE

5 letter name

KARIN

Results

2 letter name

BO OB

For this name, there are only 2 possible combinations

3 letter name

SAM ASM MSA

SMA AMS MAS

For the 3 letter name, there are 6 possible ways.

4 letter name

MIKE IMKE KMIE EMIK

MIEK IMEK KMEI EMKI

MKIE IEMK KIEM EIKM

MKEI IEKM KIME EIMK

MEIK IKEM KEIM EKMI

MEKI IKME KEMI EKIM

24 different combinations are possible with a 4 letter name

5 letter name

KARIN KRAIN KINAR KNARI

KARNI KRANI KINRA KNAIR

KAINR KRNIA KIARN KNRAI

KAIRN KRNAI KIANR KNRIA

KANRI KRINA KIRAN KNIRA

KANIR KRIAN KIRNA KNIAR

ARINK AINKR ANKRI AKRIN

ARIKN AINRK ANKIR AKRNI

ARKIN AIKRN ANIKR AKRNI

ARKNI AIKNR ANIRK AKINR

ARNKI AIRKN ANRKI AKNIR

ARNIK AIRNK ANRIK AKNRI

RINKA RNKAI RKAIN RAKIN

RINAK RNKIA RKANI RAKNI

RIKAN RNAKI RKINA RANIK

RIKNA RNAIK RKIAN RANKI

RIAKN RNIAK RKNIA RAIKN

RIANK RNIKA RKNAI RAINK

NIKAR NKIAR NAIRK NRAKI

NIKRA NKIRA NAIKR NRAIK

NIARK NKARI NARIK NRKAI

Middle

2 letter name

AA

3 letter name

AAL

4 letter name

ALAE

5 letter name

ANNIE

Also, I will look at the name Emma as part of our investigation, and to help prove my results and formula correct.

RESULTS

2 letter name

AA

For this, there is only 1 possible combination.

3 letter name

AAL

LAA

ALA

3 possible combinations are possible with this name

4 letter name

ALAE LAEA EAAL

ALEA LAAE EALA

AAEL LEAA ELAA

AALE

AELA

AEAL

For a 4 letter name, there are 12 different combinations

5 letter name

ANNIE ENNIA INNAE NNAEI NEANI

ANNEI ENNAI INNEA NNAIE NEAIN

ANEIN ENAIN INENA NNEAI NEIAN

ANIEN ENANI INEAN NNEIA NEINA

ANENI ENINA INANE NNIAE NENAI

ANIEN ENIAN INAEN NNIEA NENIA

AENIN EANNI IENNA NIEAN NAINE

AEINN EANIN IENAN NIENE NAIEN

AENNI EAINN IEANN NINAE NAENI

AINEN EINNA IANNE NINEA NAEIN

AINNE EIANN IAENN NIAEN NANIE

AIENN EINAN IANEN NIANE NANEI

For a 5 letter name there are 60 possible combinations.

RESULTS

For names with 2 letters the same, I have done some examples, and am now going to show my results.

Table of Results

Number of letters the same | Letters in word | Combinations |

2 | 2 | 1 |

2 | 3 | 3 |

2 | 4 | 12 |

2 | 5 | 60 |

Graph of Results

FORMULA

I have now worked out a formula and will now explain it:

For a 2 letter word: 2x1=1

2

For a 3 letter word: 3x2x1= 3

2

For a 4 letter word: 4x3x2x1=12

2

For a 5 letter word:

Conclusion

RESULTS

Here are my results

Table of Results

Letters in Word | Words the same | Possible Combinations |

3 | 3 | 1 |

4 | 3 | 4 |

5 | 3 | 20 |

Graph of Results

FORMULA

I have looked at my results, and am now able to work out a formula.

For a 4 letter word: 4x3x2x1=4

3x2x1

For a 5 letter word: 5x4x3x2x1= 20

3x2x1

These results are correct, so therefore my formula is correct. I will now write out this formula in algebraic form.

C=N!/S!

C= Combinations

N= Number of letters

S = Same letters

! = Factorise

Predictions

Based on my formula, I will now make a few predictions

Letters in Word | Words the Same | Possible Combinations |

6 | 3 | 120 |

7 | 3 | 840 |

8 | 3 | 6720 |

9 | 3 | 60480 |

10 | 3 | 604800 |

The reason we divide it by 3 factorised, is because there are 3 letters the same, and if we didn’t divide it, it would give the results as if each same letter was different. This means we must divide it by 3 factorised to get the right result.

Names with 4 letters the same

Now I will look at names with 4 letters the same. I will use the following names

4 letter name

AAAA

5 letter name

AAAAL

RESULTS

4 letter word

AAAA

Only 1 possible combination

5 letter word

AAAAL

LAAAA

ALAAA

AALAA

AAALA

5 possible combinations

RESULTS

Table of Results

Number of letters | Letters the same | Combinations |

4 | 4 | 1 |

5 | 4 | 5 |

Graph of Results

FORMULA

Now I will attempt to find a formula based on my results

For a 5 letter word: 5x4x3x2x1=5

4x3x2x1

This is correct, which proves that my results were correct. I shall now show the algebraic formula, which is

C=N!/S!

C= Combinations

N= Number of letters

S = Same letters

! = Factorise

PREDICTIONS

Number of Letters | Letters the Same | Combinations |

6 | 4 | 30 |

7 | 4 | 210 |

8 | 4 | 1680 |

9 | 4 | 15120 |

10 | 4 | 151200 |

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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