• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  • Level: GCSE
  • Subject: Maths
  • Word count: 2463

Emma's dilemma

Extracts from this document...

Introduction

Emma’s dilemma

image00.png

THAMILINI KUGARAJAH

CATERHAM HIGH SCHOOLimage11.png

image12.pngimage19.png

image29.png

image19.png

CENTER NO: 13321

image01.png

image01.png

CANDIDATE NO: 9191

image01.png

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.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Emma's Dilemma

    This is due to the fact that the six letters in the previous number of arrangements can be rearranged with another letter. This increases the total number of arrangements , for this number of arrangements, by a factor of 7.

  2. I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

    and as the letter B is repeated 2 times in the word we need to use 2! The formula is therefore: Y = 6! / 4! x 2! Y = 720 / 24 x 2 Y = 720 / 48 Y = 15 This produces the number 15 which is the correct amount of different arrangements.

  1. EMMA'S DILEMMA

    1*2*3*4 1*2 = 12 So the formula does not work. But f I Now lets try a 6-lettered word, which 1 letter is repeated 3 times and the letter is also repeated 3 times: 1. AAABBB 11. BBBAAA 2. AABABB 12.

  2. Emma's Dilemma

    and two repeats, so I will plot all the results I have obtained so far in a table. With these results I will try to work out a pattern and fill in the gap. This table of discoveries is shown below: Five Lettered Word With X Number Of Repeats Number

  1. Emma's Dilemma

    Number of letters Double letter, double letter. Double letter, triple letter. Triple letter, triple letter. 4 6 5 30 10 6 180 60 20 The numbers in blue are the number of arrangements that I have predicted using the formula. These should be accurate if my theory is correct.

  2. Emma's Dilemma

    It may have applied for "Emma" where 24 divided by 2 is undoubtedly 12 but it doesn't apply for the word "baaa" - 24 divided by 3 isn't 4 - or "oooo" either - 24 divided by 4 isn't 1.

  1. Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

    1, timed by the number of letters minus 2, and so on, until the number is 1 itself. No. of = No. of letters X ( No. of letters - 1 ) X ( No. of letters - 2 )

  2. We are investigating the number of different arrangements of letters.

    are only two xi need to mutiply, and if there are two pair of different number of figures of same number, then there would be x1i and x2i need to mutiply. Let's confirm the formular. let's try 112233, according the formula, I expect the total arrangements is a=(1*2*3*4*5*6)/(1*2*1*2*1*2)=90 Let's confirmed

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work