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Emma's Dilemma

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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)MEMAimage00.png

  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)KANimage08.pngimage10.png

2)KNA

3)NKA

4)NAK

5)AKN

6)ANK

...read more.

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)FFFFKimage01.png

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

...read more.

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=120image05.png

(1x2x3)x(1x2)=12image06.png

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)

image07.png

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)=240image09.png

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                                

...read more.

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