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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 = 6

The total rearrangements for words with 3 letters, 2 of them being the same letter = 3

If 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

...read more.

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

...read more.

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

image06.png

2x1=2

3x2=6                

4x6=24image01.png

                                                Using the explanation in these paragraphs

I made the formula.

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.

                Page  of

...read more.

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