# Emma's Dilemma

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Introduction

GCSE Coursework –

Emma’s Dilemma!

I am going to investigate the number of different arrangements of the letters of Emma’s name. I will investigate this by using a system. My system is to find all the arrangements beginning with ‘E’, then all the arrangements beginning with ‘M’, and finally all the arrangements beginning with ‘A’.

EMMA | MMAE | AEMM |

EMAM | MMEA | AMME |

EAMM | MAME | AMEM |

MEMA | ||

MEAM | ||

MAEM |

I have found 12 different arrangements for the name Emma. I have also found a new systematic way, to find the arrangements. This system is to find everything beginning with the same letter, by fixing two, moving two and so on.

I will use this to investigate the number of different arrangements for the name Lucy.

LUCY | UCYL | CYLU | YLUC |

LUYC | UCLY | CYUL | YLCU |

LCYU | UYLC | CLUY | YUCL |

LCUY | UYCL | CLYU | YULC |

LYUC | ULCY | CUYL | YCLU |

LYCU | ULYC | CULY | YCUL |

I have found 24 different arrangements for the name Lucy.

So far I have used names with 4 letters but ‘Emma’ has a double letter and ‘Lucy’ does not. So this means a double letter equals half the number of arrangements of a name without single letters. You can use anything to work this out – symbols, numbers etc. and the number of combinations will still be the same.

But if you were to make the double letters different I predict that there will be the same amount of arrangements as Lucy’s name – 24. I will now try to prove this:

EMMA | MMAE | MMAE | AEMM |

EMMA | MMEA | MMEA | AEMM |

EMAM | MAEM | MAEM | AMME |

EMAM | MAME | MAME | AMEM |

EAMM | MEMA | MEMA | AMEM |

EAMM | MEAM | MEAM | AMME |

Each of these arrangements are in pairs.

Middle

RSOIE

OISER

SEIRO

IREOS

EORSI

REOIS

ORSEI

SOIRE

ISEOR

EIRSO

REOSI

ORSIE

SOIER

ISERO

EIROS

REISO

OREIS

SOREI

ISORE

EISOR

REIOS

ORESI

SORIE

ISOER

EISRO

RESOI

ORISE

SOEIR

ISROE

EIORS

RESIO

ORIES

SOERI

ISREO

EIOSR

RISEO

OEIRS

SREOI

IORES

ESOIR

RISOE

OEISR

SREIO

IORSE

ESORI

RIEOS

OERSI

SROIE

IOESR

ESIRO

RIESO

OERIS

SROEI

IOERS

ESIOR

RIOSE

OESIR

SRIEO

IOSRE

ESROI

RIOES

OESRI

SRIOE

IOSER

ESRIO

I have found 120 different arrangements for the name Rosie.

I have also proved my prediction to be right.

You can also predict this in a different way. We know how many combinations are in a four-letter word, so you can also multiply the number of letters by the arrangements for a four-letter word. For example GEMMA, we know EMMA has 12 arrangements and if we multiply it by 5 we will get: 12*5=60, which is the correct number of arrangements.

I will now investigate the name Tasha.

TASHA | ASHAT | SHATA | HATAS |

TASAH | ASHTA | SHAAT | HATSA |

TAHAS | ASATH | SHTAA | HAAST |

TAHSA | ASAHT | SATAH | HAATS |

TAASH | ASTHA | SATHA | HASTA |

TAAHS | ASTAH | SAAHT | HASAT |

TSAHA | AHTAS | SAATH | HTSAA |

TSHAA | AHTSA | SAHAT | HTASA |

TSAAH | AHAST | SAHTA | HTAAS |

THAAS | AHATS | STHAA | HSAAT |

THASA | AHSTA | STAAH | HSATA |

THSAA | AHSAT | STAHA | HSTAA |

ATSAH | |||

ATSHA | |||

ATAHS | |||

ATASH | |||

ATHSA | |||

ATHAS | |||

AAHTS | |||

AAHST | |||

AATSH | |||

AATHS | |||

AASHT | |||

AASTH |

I have found 60 arrangements for the name Tasha.

If I made the make the ‘A’s different there would be the same amount of arrangements as Rosie.

Table of Results.

Number of arrangements with: | ||

Number of letters | Double letters | Single letters |

1 | 1 | |

2 | 1 | 2 |

3 | 3 | 6 |

4 | 12 | 24 |

5 | 60 | 120 |

I have now seen what the formula is. The formula that can be used is the factorial formula. To find out the number of single letters you use x! (X = the number of letters in the name), to find out the number of double letters you use (x!) / 2 (x = the number of letters in the name). This is because to get the number of arrangements for a three-letter name you have to multiply the one-letter arrangement by two, then you have to multiply it by three. These numbers are known as factorial numbers, the pattern goes like this:

1! = 1

2! = 1 x 2 =2

3! = 1 x 2 x 3 = 6

4! = 1 x 2 x 3 x 4 = 24 … etc.

We can also work out the nth term using this:

N! = 1*2*3*4*5……….(n–1)*n.

This works because (4-1)*4 = 12 and this is equal to 4! /2 is.

Using this formula I can find the number of arrangements for a six-letter name with a double letter and a six-letter name with single letters. To find out the double letter arrangement I would use (6!) / 2 which is 360, and the single letter arrangement is 6! which is 720.

I predict that if you have a triple letter in a name the number of arrangements will be a third of the double letter arrangement. Or to find the number of arrangements you could divide the single letter arrangements by 6.

This is what I think the results may be if a name had a triple letter:

Number of letters | No Repeats | Double letter | Triple letter |

3 | 6 | 3 | 1 |

4 | 24 | 12 | 4 |

5 | 120 | 60 | 20 |

To seeif my predictions are correct, I will investigate names with a triple letter.

The first name I will use is a three-letter name –XXX

XXX |

There is only one arrangement for XXX.

I will now investigate a four-letter word, with a triple letter; this name will be Lill.

LILL | ILLL |

LLLI | |

LLIL |

Conclusion

Double letter, double letter = (4! / 2!) / 2! = 6

Double letter, triple letter = (5! / 3!) / 2! = 10

Triple letter, triple letter = (6! / 3!) / 3! = 20

These are the same as my results so the formula is correct.

I will now prove that N! works

X!Y!

XXAA has 6 arrangements.

4! = 24 = 24 = 6

2!2! 2 x 2 4

XXXAA has 10 arrangements.

5! = 120 = 120 = 10

3!2! 6 x 2 12

XXXAAA has 20 arrrangements.

6! = 720 = 720 = 20

3!3! 6 x 6 36

In conclusion to my investigation, I have found out that a name with a double letter has half the arrangements of a name (with the same number of letters) with no repeated letters. A name with 3-letters the same has a 6th of the arrangements of a name with no repeats and a 3rd of a name with a double letter. I have also found out that a name with four letters the same has a 24th of the arrangements of a name with no repeats and a 4th of a name with a triple letter.

I have seen how the factorial numbers make it easier to work out the amount of arrangements a word has. I have also found which formula works to find out the number of arrangements names with double letter-double letter, double letter-triple letter and triple letter-triple letter. This formula is

N!

X!Y!

I have also found out that a five-letter word with all the letters different has the same amount of arrangements as a four-letter word multiplied by five. The formula would be 4! * 5, the overall formula is (N-1)*N

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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