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• Level: GCSE
• Subject: Maths
• Word count: 1872

# Emma's Dilemma

Extracts from this document...

Introduction

Nasim A.A        Word Rearrangements        Maths Coursework

 1 Liu Ted Sam aab 2 Lui Tde Sma aba 3 Uli Det Mas baa 4 Uil Dte Msa 5 Ilu Etd Asm 6 Iul Edt Ams Total rearrangements for words with 3 letters, none of them being the same letter = 6The total rearrangements for words with 3 letters, 2 of them being the same letter = 3If a word has three letters, all of them being the same letter, it can not be rearranged.

To start this project I am going to write my prediction, followed by the rearrangements of some words.  Later I will compare the results and find out if there is a connection between the numbers of rearrangements for different words.

I predict that words that have the same number of letters, will have the same number of rearrangements.  This is because the words will only have different letters, but the same rearrangements, i.e. I could replace all of the letters in a word with a, b, c etc.

 1 Lucy Mary John 2 Luyc Mayr Jonh 3 Lyuc Myar Jnoh 4 Lycu Myra Jnho 5 Lcuy Mrya Jhon 6 Lcyu Mray Jhno 7 Ulcy Amyr Ojnh 8 Ulyc Amry Ojhn 9 Uylc Army Ohnj 10 Uycl Arym Ohjn 11 Ucyl Ayrm Onjh 12 Ucly Aymr Onhj 13 Culy Ramy Honj 14 Cuyl Raym Hojn 15 Cylu Ryam Hjon 16 Cyul Ryma Hjno 17 Cluy Rmya Hnoj 18 Clyu Rmay Hnjo 19 Yclu Yamr Nojh 20 Ycul Yarm Nohj 21 Yucl Ymar Nhjo 22 Yulc Ymra Nhoj 23 Ylcu Yram Njoh 24 Yluc Yrma Njho Total rearrangements for words with 4 letters, none of them being the same letter = 24
 1 Emma Jazz 2 Emam Jzaz 3 Eamm Jzza 4 Amme Ajzz 5 Amem Azjz 6 Aemm Azzj 7 Mmae Zzja 8 Mmea Zzaj 9 Maem Zazj 10 Mame Zajz 11 Mema Zjaz 12 Meam Zjza

Middle

Slhae

Alhes

Hsael

Ealsh

Lhsae

18

Slhea

Alhse

Hsale

Ealhs

Lhsea

19

Sheal

Aelsh

Hlsae

Elhas

Leash

20

Shela

Aelhs

Hlsea

Elhsa

Leahs

21

Shlae

Aehls

Hlesa

Elsha

Lehsa

22

Shlea

Aehsl

Hleas

Elsah

Lehas

23

Shale

Aeshl

Hlaes

Elash

Lesah

24

Shael

Aeslh

Hlasa

Elahs

Lesha

1

Nasim

Ansim

Smina

Imnas

Masin

2

Nasmi

Ansmi

Smian

Imnsa

Masni

3

Namsi

Anmsi

Smain

Imsan

Mansi

4

Namis

Anmis

Smani

Imsna

Manis

5

Naism

Anism

Smnia

Imasn

Maisn

6

Naims

Anims

Smnai

Imans

Mains

7

Nsima

Amnis

Snmai

Iamns

Msina

8

Nsiam

Amnsi

Snmia

Iamsn

Msian

9

Nsami

Amsin

Sniam

Iasnm

Msain

10

Nsaim

Amsni

Snima

Iasmn

Msani

11

Nsmai

Amins

Snami

Iansm

Msnia

12

Nsmia

Amisn

Snaim

Ianms

Msnai

13

Niasm

Asmin

Samin

Insam

Minsa

14

Niams

Asmni

Samni

Insma

Minas

15

Nimas

Asnim

Sanim

Inmas

Mians

16

Nimsa

Asnmi

Sanmi

Inmsa

Miasn

17

Nisam

Asimn

Sainm

Inasm

Misan

18

Nisma

Asinm

Saimn

Inams

Misna

19

Nmisa

Aimsn

Siamn

Isnam

Mnisa

20

Nmias

Aimns

Sianm

Isnma

Mnias

21

Nmsia

Ainms

Sinam

Ismna

Mnsai

22

Nmsai

Ainsm

Sinma

Isman

Mnsia

23

Nmasi

Aismn

Simna

Isanm

Mnasi

24

Nmais

Aisnm

Siman

Isamn

Mnais

Total rearrangements for names with 5 letters, none of them being the same = 120

 Criss Isscr Scris Crsis Issrc Scrsi Crssi Isrsc Scsri Csris Isrcs Scsir Csrsi Iscrs Scirs Cssri Iscsr Scisr Cssir Icrss Sicrs Csirs Icsrs Sicsr Csisr Icssr Siscr Cissr Ircss Sisrc Cisrs Irscs Sircs Cirss Irssc Sirsc Rciss Sscri Rcsis Sscir Rcssi Ssicr Rssci Ssirc Rssic Ssric Rsisc Ssrci Rsics Srsci Rscis Srsic Rscsi Srisc Ricss Srics Riscs Srcis Rissc Srcsi

The total rearrangements for any word with 5 letters, 2 of them being the same letter = 60

 1 Abccc 2 Acbcc 3 Accbc 4 Acccb 5 Baccc 6

Conclusion

a=number of same letter(1) in a word

b=number of same letter(2) in a word

c=number of same letter(3) in a word

From this information I can make some new formulas for different words:

Aabbcc        This has the formula, ‘6!/2!x2!x2!’ which equals 90 rearrangements

Aaabbbcc        This has the formula, ‘11!/3!x3!x2!’ which equals 554400 rearrangements

Conclusion

My prediction was incorrect!  What I did to begin with was:

A1A2BB

A2A1BB

I did not know that this was not permitted until I was told so by person X.

Evaluation

To find my first formula, ‘n(n-1)(n-2)(n-3)…(n-(n-1))’, I used the fact that the number of letters in that word was multiplied by the number of rearrangements for the previous number;

 1 letters the same No. letters Rearrang-ements 1 1 2 2 3 6 4 24 5 120

2x1=2

3x2=6

4x6=24

Using the explanation in these paragraphs

The reason I always put a ‘1’ at the end of each equation (such as 3x2x1) is because I found it difficult to put an ending to my first formula.  Without a 1, my formula would be ongoing, i.e.

N(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)…

I made a mistake, forgetting about words that contain two groups of letters.  This put me behind.  If I had remembered, I could have saved myself some time.

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