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  • Level: GCSE
  • Subject: Maths
  • Word count: 2463

Emma's dilemma

Extracts from this document...

Introduction

Emma’s dilemma

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THAMILINI KUGARAJAH

CATERHAM HIGH SCHOOLimage11.png

image12.pngimage19.png

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CENTER NO: 13321

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CANDIDATE NO: 9191

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Introduction:

This investigation is about finding the different arrangements of letters in various words. I am going to look at the number of different arrangements for words with all letters different, words with two letters the same  and the words with three letters the same and so on.

  • Part 1

I am going to look at the number of different arrangements of the letters of Lucy’s name.

LUCY

  1. LUCY
  2. LUYC
  3. LCYU
  4. LCUY
  5. LYUC
  6. LYLU
  1. ULYC
  2. ULCY
  3. UCLY
  4. UCYL
  5. UYCL
  6. UYLC
  1. CYLU
  2. CYUL
  3. CULY
  4. CUYL
  5. CLYU
  6. CLUY
  1. YLUC
  2. YLCU
  3. YUCL
  4. YULC
  5. YULU
  6. YCUL

As it shown above, there are 24 different arrangements in four letter word.

CYimage03.png

        Uimage04.png

YCimage06.pngimage05.png

UYimage07.pngimage03.png

L        Cimage08.pngimage08.png

YUimage09.pngimage05.png

 CUimage05.png

Yimage10.png

U        Cimage05.png

A tree diagram can also be used as I have used above. Here we have used “L” as a first letter; however the same process can be used with other 3 letter, so we can multiply the number of letter by the number of possibilities, 6*4=24

 Each starting letter has six possibilities. Therefore four letters altogether 6+6+6+6=24

You will get 24 possibilities altogether.

First of all we have got 4 possibilities (4 letters) and then 3 possibilities(by leaving first letter “L”) and then 2 possibilities 9by leaving first two letters “L”U”) after that one possibility which is the last letter . Therefore 4*3*2*1=4! Which is equal to 24, where as we can see this from tree diagram (which is shown by using different colours)

  • Part 2

I am going to investigate the number of different arrangements of the letters of EMMA’S name.

EMMA

  1. EMMA
  2. EMAM
  3. EAMM
  4. AMEM
  5. AEMM
  6. AMME
  1. MAEM
  2. MEAM
  3. MEMA
  4. MAME
  5. MMAE
  6. MMEA

...read more.

Middle

EBC
  1. AEBCD
  2. AEBDC
  3. AEDBC
  4. AEDCB
  5. AECDB
  6. AECBD

There are 24 arrangements starting just with one letter “A”. therefore there are five letters altogether, so 5*24=120

Consequently there are 120 arrangements altogether.

51 = 5*4*3*2*1

= 120 arrangements.

Now I am going to draw a table of results and see whoat relationship between them.

Table 1

Number of letters in the word

Number of different arrangements

observation

1

1

1=1!

2

2

1*2=2!

3

6

1*2*3=3!

4image18.png

24

1*2*3*4=4!

5image20.png

120

1*2*3*4*5=5!

image21.png

                              5*24   =120                (cross multiple)

I have noticed that for the word with all letters different, the factorial of the number of the letters in the word is equal to the number of different arrangements.

Let “N”=number of letters in the word

“A”= number of different arrangements

Therefore N! =A

For example: a three letter word with all letters different

Prediction: for the word with six letters different.

Using pattern noticed

1)          N*(N-1) =A

6*5! =6*120

        =720

Using the formula

  1. A=N!

A=6!

A=720

The calculations using formula and the pattern noticed both agree.

I am going to investigate words with two letters the same.

  1. HH – this has one arrangement
  1.  TOO

        OTO        -this has three arrangements

        OOT

  1. A word with four letters with two letters the same will give the same number of arrangements as EMMA i.e. 24 arrangements
  1. The arrangement for a word with five letters the two letters the same is shown on the next page. I have used the word “PAPER” and I have numbered each of the letter “p “to make them different from each other.

This is to enable me to work out the total number of arrangements when all the letters are different and to compare it when two of the letters are the same.        

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

  • PAPER
  • PAEPR
  • PAERP
  • PAPRE
  • PEREP
  • PARPE
  • PPAER
  • PPEAR
  • PPERA
  • PPARE
  • PPREA
  • PPRAE
  • PEAPR
  • PEARP
  • PEARP
  • PERPA
  • PEPRA
  • PEPAR
  • PRAEP
  • PRAPE
  • PREAP
  • PREPA
  • PRPEA
  • PRPAE
  • RAPPE
  • RAEPP
  • RAPEP
  • ARPEP
  • ARPPE
  • AREPP
  • PAPER
  • PAEPR
  • PAERP
  • PAPRE
  • PEREP
  • PARPE
  • PPAER
  • PPEAR
  • PPERA
  • PPARE
  • PPREA
  • PPRAE
  • PEAPR
  • PEARP
  • PEARP
  • PERPA
  • PEPRA
  • PEPAR
  • PRAEP
  • PRAPE
  • PREAP
  • PREPA
  • PRPEA
  • PRPAE
  • RAPPE
  • RAEPP
  • RAPEP
  • ARPEP
  • ARPPE
  • AREPP

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

  • APPER
  • APPRE
  • APREP
  • APRPE
  • APEPR
  • AEPRP
  • AEPPR
  • AERPP
  • REPAP
  • REPAP
  • REEPPA
  • REAPP
  • RPPEA
  • RPPAE
  • RPEAP
  • RPEPA
  • RPAEP
  • RPAPE
  • EAPRA
  • EAPPR
  • EARPP
  • ERPAP
  • ERPPA
  • ARAPP
  • EPPAR
  • EPPRA
  • EPARP
  • EPAPR
  • EPRAP
  • EPRPA
  • APPER
  • APPRE
  • APREP
  • APRPE
  • APEPR
  • AEPRP
  • AEPPR
  • AERPP
  • REPAP
  • REPAP
  • REPPA
  • REAPP
  • RPPEA
  • RPPAE
  • RPEAP
  • RPEPA
  • RPAEP
  • RPAPE
  • EAPRA
  • EAPPR
  • EARPP
  • ERPAP
  • ERPPA
  • ARAPP
  • EPPAR
  • EPPRA
  • EPARP
  • EPAPR
  • EPRAP
  • EPRPA

There are 60 different arrangements altogether, however

PAPERimage22.png

PAPER        1 arrangement

We can see two arrangements, when we swap that “P”,”P” but as they look the same there are 120 arrangements when we use the P as P AND P. however as they look the same we divide 120 by two  which is equal to sixty arrangements.

Now I am going to put the results on the table and see what relationships between them.

The word with two letters the same

Table 2

Number of letters in the word

Number of different arrangements

observation

2

1

2!/2

3

3

3!/2

4image23.png

12

4!/2

5image24.png

60image25.png

5!/2

...read more.

Conclusion

Therefore N! /R! =A

Prediction:

Testing the formula

N! R! =A

e.g. =MUMMY

=5!/3!=120/6=20 different arrangements

This proves my formula is right and it would work for the word that has any number of letters the same.

Multiple number of letters the same

Now I am going to look at the group of letters that has more than one letters the same, example: “ABBCC, AABBCC,” etc

 First of all I am going to look at “AABB”

  1. AABB
  2. ABAB
  3. BABA
  4. BBAA
  5. BAAB
  6. ABBA

There are six different arrangements.

I know numerator should be N!. I know this from the previous section.

From this section I have noticed that for the word with multiple number of letters the same..

The number of different arrangements is equal to the factorial of the different no of multiple letters the same.

Example: there are four letters in the word and two different the same

So 4! /2!*2!

Let “N”=number of letters in the word

“A”= number of different arrangements

R

Q        =number of letters the same.

p        different number of letters appears this number  of time e.g. P times.

In a “N” letter word same letters appears “P, Q, R” times. So the number of ways we could arrange is

N! /P!*Q!*R! =A

Conclusion

In this investigation, I have learned how to find the different number of arrangements for the word with any of different letters or the word with different number of letters the same.

I found three different formulae:

  • Word with different letters                                        N! /A
  • Words with more than one letter the same                        N! /R! =A
  • Words with multiple letters the same                                 N! /R!*P! =A

...read more.

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