# 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.

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

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.

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

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