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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.

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

(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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