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  • Level: GCSE
  • Subject: Maths
  • Word count: 1561

Emma's Dillemma

Extracts from this document...

Introduction

1.

1

Emma

2

Emam

3

Eamm

4

Aemm

5

Amem

6

Amme

7

Meam

8

Mema

9

Mame

10

Maem

11

Mmae

12

Mmea

There are 12 different arrangements of Emma

2.

1

Lucy

2

Luyc

3

Lyuc

4

Lycu

5

Lcuy

6

Lcyu

7

Culy

8

Cuyl

9

Cylu

10

Cyul

11

Cluy

12

Clyu

13

Yclu

14

Ycul

15

Ylcu

16

Yluc

17

Yucl

18

Yulc

19

Ulcy

20

Ulyc

21

Uylc

22

Uycl

23

Uclu

24

Ucul

There are 24 different arrangements of Lucy

3.

1

Jo

2

Oj

1

Joe

2

Jeo

3

Eoj

4

Ejo

5

Oje

6

Oej

1

Ann

2

Nan

3

Nna

1

JJ

Number of Letters

With Repeated Letters

No Repeated Letters

JJ

Jo

2

1

2

Ann

Joe

3

3

6

Emma

Lucy

4

12

24

The “no repeated letters” field is equal to the “With repeated letters” field but doubled i.e

Lucy = 4 letters = 24 arrangements all you need to do then is to divide it by 2 and you will have the value of “With repeated letters” field and visa versa.

There is a quicker solution to this problem. All that you need to do is (If you need a 4 letter word) you would do 1x2x3x4. If you needed 5 letters you would do 1x2x3x4x5 and so on. If you want to find out the value of the repeated letters field you would then just divide it by 2.

I predict that a 5 letter word will have 120 combinations because 1x2x3x4x5 = 120.

...read more.

Middle

1

Hannah

31

Ahnnha

61

Nhaahn

2

Hannha

32

Ahnnah

62

Nhaanh

3

Hanhna

33

Ahnanh

63

Nhanah

4

Hanhan

34

Ahnahn

64

Nhanha

5

Hanahn

35

Ahnhan

65

Nhahna

6

Hananh

36

Ahnhna

66

Nhahan

7

Haannh

37

Ahhnna

67

Nhhaan

8

Haanhn

38

Ahhnan

68

Nhhana

9

Haahnn

39

Ahhann

69

Nhhnaa

10

Hahnan

40

Ahanhn

70

Nhnaha

11

Hahnna

41

Ahannh

71

Nhnaah

12

Hahann

42

Ahahnn

72

Nhnhaa

13

Hhanna

43

Aahnnh

73

Nnhaah

14

Hhanan

44

Aahnhn

74

Nnhaha

15

Hhaann

45

Aahhnn

75

Nnhhaa

16

Hhnaan

46

Aanhhn

76

Nnahha

17

Hhnana

47

Aanhnh

77

Nnahah

18

Hhnnaa

48

Aannhh

78

Nnaahh

19

Hnnaah

49

Annhha

79

Naahhn

20

Hnnaha

50

Annhah

80

Naahnh

21

Hnnhaa

51

Annahh

81

Naanhh

22

Hnanha

52

Anhnah

82

Nahanh

23

Hnanah

53

Anhnha

83

Nahahn

24

Hnhaan

54

Anahhn

84

Nanhha

25

Hnhana

55

Anahnh

85

Nanhah

26

Hnhnaa

56

Ananhh

86

Nanahh

27

Hnahna

57

Anhanh

87

Nahnah

28

Hnahan

58

Anhahn

88

Nahnha

29

Hnaahn

59

Anhhan

89

Nahhna

30

Hnaanh

60

Anhhna

90

Nahhan

Hannah

If a word contains six letters – all of which are doubles e.g. Hannah, it would be able to be rearranged 90 times (according to my theory, which would be (1x2x3x4x5x6)

(1x2)(1x2)(1x2)  = 90

Why do we halve?

Proof

1

emMa

7

amMe

13

Mmae

19

mMae

2

emaM

8

ameM

14

Mmea

20

mMea

3

eMma

9

aMme

15

Mame

21

maMe

4

eMam

10

aMem

16

Maem

22

maeM

5

eaMm

11

aeMm

17

Meam

23

meaM

6

eamM

12

aemM

18

Mema

24

meMa

If you treat the double letters as individual letters then you don’t needto halve in the formula. You halve because of the double letter, the example above proves it.

Factorial – A shorthand algebraic method

...read more.

Conclusion

3

YYXXY

8

XYXYY

4

YYXYX

9

XYYXY

5

YXXYY

10

XYYYX

Four X’s and Four Y’s

A double quad should add up to 70 because  8!

                                                               (4!)(4!)

Proof

1

XXXXYYYY

36

YYYYXXXX

2

XXXYYYYX

37

YYYXXXXY

3

XXXYYYXY

38

YYYXXXYX

4

XXXYYXYY

39

YYYXXYXX

5

XXXYXYYY

40

YYYXYXXX

6

XXYYYYXX

41

YYXXXXYY

7

XXYYXXYY

42

YYXXYYXX

8

XXYXXYYY

43

YYXYYXXX

9

XXYXYXXY

44

YYXYXYYX

10

XXYXYXYX

45

YYXYXYXY

11

XXYXYYXX

46

YYXYXXYY

12

XXYYXYXY

47

YYXXYXYX

13

XXYYXYYX

48

YYXXYXXY

14

XYXXXYYY

49

YXYYYXXX

15

XYXXYYYX

50

YXYYXXXY

16

XYXXYYXY

51

YXYYXXYX

17

XYXXYXYY

52

YXYYXYXX

18

XYYXXXYY

53

YXXYYYXX

19

XYYXXYXY

54

YXXYYXYX

20

XYYXXYYX

55

YXXYYXXY

21

XYYXYXXY

56

YXXYXYYX

22

XYYYXXXY

57

YXXXYYYX

23

XYYYXXYX

58

YXXXYYXY

24

XYYYXYXX

59

YXXXYXYY

25

XYYYYXXX

60

YXXXXYYY

26

XYXYXYXY

61

YXYXYXYX

27

XYXXXXYY

62

YXYYYYXX

28

XYXYXYYX

63

YXYXYXXY

29

XYXYYXXY

64

YXYXXYYX

30

XYXYYXYX

65

YXYXXYXY

31

XXYXYXYY

66

YYXYXYXX

32

XXYXYXXY

67

YYXYXYYX

33

XXYXYXYX

68

YYXYXYXY

34

XXXYXYYY

69

YYYXYXXX

35

XXXYYXYY

70

YYYXXYXX

General Equation

To find a general equation I will make a table of results:

Arrangements

Number of letters

Equation

Repeted letters

Answer

XX

2

2!

2

1

(2!)

XXY

3

3!

2

3

2!

XXYY

4

4!

2+2

6

(2!)(2!)

XXXY

4

4!

3

6

3!

XXXXYYYY

8

8!

4+4

7

(4!)(4!)

XXXXYY

6

6!

4+2

15

(4!)(2!)

As you can see by the table there is an undisputed relationship between the number of repeted letters, the number of letters and the formula.

General Formula

Using the table of results I can make a formula which will fit every word :

(X+Y)!

(X!)+(Y!)  

Conclusion

To conclude, any word no matter how long or haw many repeats it has can be solved using the above formula for example Computation has eleven letters in it and the letter o is  repeted twice, t is repeted twice, c is repeted twice and n is repeted twice so the formula for it would be:

11!

(2!)(2!)(2!)(2!)

That would equal 2494800 combinations.

...read more.

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