# Investigate the number of arrangements of Dave's name.........

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Introduction

Maths Coursework

1. Investigate the number of arrangements of Dave’s name………

DAVE

DAEV

DEAV

DEVA

DVAE

DVEA

ADEV

ADVE

AEDV

AEVD

AVED

AVDE

VDAE

VDEA

VEDA

VEAD

VAED

VADE

EVDA

EVAD

EDVA

EDAV

EAVD

EADV

There are 24 different arrangements for Dave’s name.

- For a two letter word (AT) , there are two arrangements.

AT

TA

- For a three letter(CAT) word, there are six arrangements.

CAT

CTA

ACT

ATC

TAC

TCA

- For a four letter word (LEAD), there are twenty-four arrangements.

LEAD

LEDA

LAED

LADE

LDEA

LDAE

DEAL

DELA

DLEA

DLAE

DALE

DAEL

ADEL

ADLE

ALED

ALDE

AEDL

AELD

EALD

EADL

EDAL

EDLA

ELDA

ELAD

- For a five letter word (SOUTH), there are one hundred and twenty arrangements.

SOUTH

SOTUH

SOHUT

SOUHT

SOHTU

SOTHU

SUOTH

SUOHT

SUHOT

SUHTO

SUTOH

SUTHO

STUHO

STUOH

STHUO

STHOU

STOHU

STOUH

SHTOU

SHTUO

SHUOT

SHUTO

SHOYU

SHOUT

OSUTH

OSUHT

OSHUT

OSHTU

OSTHU

OSTUH

OTSUH

OTSHU

OTHSU

OTHUS

OTUSH

OTUHS

OUTHS

OUTSH

OUSTH

OUSHT

OUHST

OUHTS

OHUTS

OHUST

OHSUT

OHSTU

OHTSU

OHTUS

UOSTH

UOSHT

UOHST

UOHTS

UOTSH

UOTHS

UTOHS

UTOSH

UTSOH

UTSHO

UTHOS

UTHSO

USTHO

USTOH

USOTH

USOHT

USHOT

USHTO

UHTSO

UHTOS

UHOTS

UHOST

UHSTO

UHSOT

THUOS

THUSO

THSUO

THOSU

THOUS

TSHOU

TSHUO

TSUHO

TSUOH

TSOUH

TSOHU

TOSHU

TOSUA

TOUSH

TOUHS

TOHSU

TOHUS

TUHSO

TUHOS

TUSOH

TUSHO

TUOHS

TUOSH

HSOUT

HSOTU

HSTUO

HSTOU

HSUTO

HSOUT

HTSOU

HTSUO

HTUSO

HTUOS

HTOUS

HTOSU

HUOST

HUOTS

HUTOS

HUTSO

HUSTO

HUSOT

HOUST

HOUTS

HOTUS

HOTSU

HOSTU

HOSUT

Middle

720

1x2x3x4x5x6

7

5040

1x2x3x4x5x6x7

From this I have concluded that the formula n! is correct( n representing the number of letters).

Therefore, to find the number of arrangements for six letter word, you would multiply the number of letters (6) by the number of arrangements of the previous number (120). This gives seven hundred and twenty arrangements.

I then tried to simplify this. By doing so I found that multiplying all the numbers below the number of letters in the given word together, the correct number of arrangements would be calculated. For example, if I was to find the number of arrangements in a seven letter word I would perform the following calculation :

1 x 2 x 3 x 4 x 5 x 6 x 7

This simplified again is 7!.

2. Investigate the number of arrangements of Emma’s name…….

EMMA

EMAM

EAMM

AEMM

AMME

AMEM

MEMA

MAEM

MMEA

MMAE

MAME

MEAM

From Emma’s name, a four letter word, we this time only get twelve arrangements. This exactly half of twenty-four,(twenty-four being the total number of arrangements for a four letter word with all the letters different, divided by two because two is the number of letters repeated). I then tried a three letter (TOO) word with two letters repeated

TOO

OTO

OOT

From TOO we get three arrangements, which is the sum of six divided by two (again, six being the number of arrangements for a three letter word with all letters different, and divide by two as it is the number of letters repeated in the word.)

From this I worked out a formula for the number of arrangements in a word which has two letters the same in it.

This was….

n!

2

This formula is demonstrated in the table below, which proves it is correct.

Number of letters | Number of letters repeated | Arrangements | Calculation |

2 | 2 | 1 | 2! divided by 2 |

3 | 2 | 3 | 3! divided by 2 |

4 | 2 | 12 | 4! divided by 2 |

5 | 2 | 60 | 5! divided by 2 |

6 | 2 | 360 | 6! divided by 2 |

7 | 2 | 2520 | 7! divided by 2 |

Conclusion

name (n!). To find a word with more than one type of letter repeated you must multiply

(y!)

both ‘y’ values together. For example………

To find a the number of arrangements of an eight letter word with two letters and two lots of three letters the same (AAABBBCC) you would calculate

8!____

(3! x 3! x 2!)

I then set about trying to find a formula for this. I quickly came to the decision that all that is needed is letters to represent the number of repeated letters. This gave me the formula:

____n!____

(x! X y! X z!)

(‘x’ represents the number of repeated A’s, ‘y’ represents the number of repeated B’s and ‘z’ represents the number of repeated C’s)

To test my formula I used a five letter word (AABBC), with two lots of letters the same. This gives me the following number of arrangements……….

CABBA

CBAAB

CAABB

CBBAA

CBABA

CABAB

AABBC

AACBB

AABCB

ABABC

ABBCA

ABCBA

ACBBA

ABBAC

ACABB

ACBAB

ABCAB

ABACB

BBAAC

BABAC

BBCAA

BAACB

BACAB

BCAAB

BAABC

BCBAA

BCABA

BACBA

BBACA

BABCA

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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