# Emma's dilemma.

Extracts from this document...

Introduction

Lauren-Faye Burgess

I am investigating the number of different combinations of letters for the name Emma.

- EMMA 12 combinations
- EMAM
- EAMM
- MEMA
- MAME
- MMAE
- MMEA
- MEAM
- MAEM
- AMME
- AMEM
- AEMM

The name Emma has 12 different combinations. As the name Emma has two M’s in it I will investigate the number of different combinations for the name Lucy.

- LUCY 24 combinations
- LUYC
- LCYU
- LCUY
- LYCU
- LYUC
- UCLY
- UCYL
- ULCY
- ULYC
- UYLC
- UYCL
- CUYL
- CULY
- CLUY
- CLYU
- CYUL
- CYLU
- YCLU
- YCUL
- YULC
- YUCL
- YLCU
- YLUC

The name Lucy has 24 combinations this is twice as many combinations as Emma this is because EMMA has two letters the same. If you were to colour code the Ms in EMMA then you may have 24 combinations. I am going to investigate this.

- EMMA
- EMMA*
- EMAM
- EMAM*
- EAMM
- EAMM*
- MEMA
- MEMA*
- MAME
- MAME*
- MMAE
- MMAE*
- MMEA
- MMEA*
- MEAM
- MEAM*
- MAEM
- MAEM*
- AMME
- AMME*
- AMEM
- AMEM*
- AEMM
- AEMM*

All of the combinations with a * next to them are cancelled out

Middle

3 letters

- Sam 6 combinations
- Sma
- Ams
- Asm
- Mas
- Msa

5 letters

- R(obin) =24 +
- O(rbin) =24 +
- B(roin) =24 +
- I(robn) =24 +
- N(robi) =24 +

=120

Here is a results table that I have put all of my results in.

Total no. of letters | No repeated letters |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

After putting my results in a table I was able to predict the number of arrangements for 5 different letters by realising a pattern. I realised that with 5 different letters there are 24 possibilities beginning with each different letter which is shown in the table where you can see that to get 120 you had to times the number of arrangements for a 4 letter word with the number of letters for a 5 letter word.

So therefore

Conclusion

Number of letters | All different | 1 letter repeated twice | 1 letter repeated 3 times | 1 letter repeated four times |

2 | 2 | 1 | ||

3 | 6 | 3 | ||

4 | 24 | 12 | 4 | 1 |

5 | 120 | 60 | 20 | 5 |

6 | 720 | 360 | 120 | 30 |

n! / 2 | n! / 6 | n! / 24 |

To make it a bit easier instead of using letters as such I will use x's and y's (any letter). I will start with xxyy:

Arrangements for xxyy:

x=a, y=b

This is a 4-letter word with 2 different. I have done this with;

aabb abab baab

aaba baba bbaa

There are 6 arrangements. What if I had xxxyy?

x=a, y=b

aaabb aabab aabba ababa abaab

abbaa bbaaa baaab babaa baaba

There are 10 different arrangements for this instance.

What if I had an arrangement of xxxxy?

x=a, y=b

aaaab aaaba aabaa

abaaa baaaa

There are 5 different arrangements for this instance.

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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