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• Level: GCSE
• Subject: Maths
• Word count: 1325

# Emma's Dilemma

Extracts from this document...

Introduction

Student name: Birkan Akin                 MathsCoursework                                              29/12/2001

## Title of coursework: Emma’s Dilemma

Aim: To investigate and find a formula for the number of different arrangements of the letters in a name. This will include names that have two, three, four or five of the same letters in a name.

In my investigation I am going to investigate the number of different arrangements of letters in a word different words.

(Task 1) I will be investigating the number of different arrangements of the letters in EMMA’s name.

1)EMMA

2)EMAM

3)EAMM

4)MEAM

5)MEMA

6)MAEM

7)MAME

8)MMEA

9)MMAE

10)AEMM

11)AMEM

12)AMME

(Task 2/3) I will be investigating the number of different arrangements of the letters in LUCY’s name, which has no letters the same. Also I will be investigating names I have chose.

1)LUCY

2)LUYC

3)LYCU

4)LYUC

5)LCYU

6)LCUY

7)ULCY

8)UCLY

9)UYLC

10)ULYC

11)UCYL

12)UYCL

13)YCUL

14)YUCL

15)YULC

16)YLCU

17)YLUC

18)YCLU

19)CYLU

20)CYUL

21)CUYL

22)CULY

23)CLUY

24)CLYU

I will now be investigating the number of different arrangements of the letters in SAM’s name.

1)KAN

2)KNA

3)NKA

4)NAK

5)AKN

6)ANK

Middle

6=

720

7=

5040

From the table of results I have found out that a 2 letter word has 2 arrangements, and a 3 letter word has 6.

Taking for example a 3 letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4!, which is called 4 factorial which is the same as 4 x 3 x 2.

So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8 letter word.

The formula for this part of the investigation is: n! = d

n = the number of letters in the word, i.e. BIRKAN has 6 letters.

d = the number of different arrangements, i.e. DAVID has 120 arrangements.

Now I am going to investigate the number of different arrangements in a word with letters repeated e.g. 3 letters repeated and 4 letters repeated.

1)LTTA

2)LTAT

3)LATT

4)TLAT

5)TLTA

6)TALT

7)TATL

8)TTLA

9)TTAL

10)ALTT

11)ATLT

12)ATTL

1)CKK

2)KKC

3)KCK

1)AQMMM

2)AMQMM

3)AMMQM

4)AMMMQ

5)QAMMM

6)QMAMM

7)QMMAM

8)QMMMA

9)MAQMM

10)MAMQM

11)MAMMQ

12)MQAMM

13)MQMAM

14)MQMMA

15)MMAQM

16)MMQMA

17)MMMAQ

18)MMMQA

19)MMAQM

20)MMMQA

1)SPPP

2)PSPP

3)PPSP

4)PPPS

1)FFFFK

2)FFFKF

3)FFKFF

4)FKFFF

5)KFFFF

I have created a result table for all ways of combinations e.g. same letter repeated.

Table of Results

 No of letters All letters different 2 letters the same 3 letters the same 4 letters the same 5 letters the same Abc 6 3 1 - - Abcd 24 12 4 1 - Abcde 120 60 20 5 1 Abcdef 720 360 120 30 6 Abcdefg 5040 2520 840 210 42

Conclusion

From this I have come up with a new formula the number of total letters factorial, divided by the number of c’s, d’s etc factorised and multiplied.

A four letter word i.e. cdcd, this has 2 c’s and 2 d’s.

Working out: n!(4)/ c!(2)/ d!(2)

1x2x3x4=24

(1x2)x(1x2)=4

24/4=6

A five letter word i.e. cccdd, this has 3 c’s and 2 d’s.

Working out: n!(5)/ c!(3)/ d!(2)

1x2x3x4x5=120

(1x2x3)x(1x2)=12

120/12=10

A six letter word i.e. ccccdd, this has 4 c’s and 2 d’s.

Working out: n!(6)/ c!(4)/ d!(2)

1x2x3x4x5x6=720

(1x2x3x4)x(1x2)=48

720/48=15

A seven letter word i.e. cccccdd, this has 5 c’s and 2 d’s.

Working out: n!(7)/ c!(5)/ d!(2)

1x2x3x4x5x6x7=5040

(1x2x3x4x5)x(1x2)=240

5040/240=21

On my previous pages I have explained, tested and showed how my formula works.

I have concluded by using examples on previous pages that Nis the number of letters in the word and that C and D is the number of repeated letters the same. The letters I have chosen i.e. Cand Dcan be different I just used those letters for a example.

Birkan Akin 11sxl

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