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# Emma&amp;#146;S Dilemma

Extracts from this document...

Introduction

Daniel Kerrigan 11.2

EMMA’S DILEMMA

I will now try some simple cases of rearranging letters to count the number of different combinations.

TWO LETTER WORDS:

AB BA

CC

THREE LETTER WORDS:

ABC  ACB  BAC  BCA  CBA  CAB

ABB  BAB  BBA

FOUR LETTER WORDS:

CLUY  CLYU  CULY  CYLU  CYUL  CUYL  LUCY  LYCU  LCYU  LCUY  LUYC  UYCL  UYLC  ULCY  ULYC  UCYL  UCLY  YCUL  YCLU  YLCU  YLUC  YUCL  YULC

AEMM  AMME  AMEM  EMMA EAMM EMAM MEMA MEAM MMEA MMAE MAME MAEM

RESULTS TABLE:

 NAME Number of different combinations CC 1 AB 2 ABB 3 ABC 6 EMMA 12 LUCY 24

## In my table I have spotted a pattern.  The pattern is when a name has two letters the same it only has half the combinations of a name with all letters different.

PREDICTION:

I predict that if a word has 2 of the same letters then it has half the number of combinations than a word with the same amount of letters, of which all are different.

Middle

ACIGR  ACIRG  ACRIG  ACRGI  ACGIR  ACGRI  AICGR  AICRG  AIGCR  AIGRC  AIRGC  AIRCG  AGIRC  AGICR  AGCIR  AGCRI  AGRCI  AGRIC  ARCIG  ARCGI  ARICG  ARIGC  ARGIC  ARGCI  CAIGR  CAIRG  CAGIR  CAGRI  CARGI  CARIG  CIARG  CIAGR  CIGAR  CIGRA  CIRGA  CIRAG  CGAIR  CGARI  CGIRA  CGIAG CGRIA  CGRAI  CRAIG  CRAGI  CRIAG  CRIGA  CRGAI  CRGIA  IACGR  IACRG  IAGRC  IAGCR  IARGC  IARCG  ICAGR  ICARG  ICGAR  ICGRA  ICRGA  ICRAG  IGARC  IGACR  IGCAR  IGCRA  IGRAC  IGRCA  IRAGC  IRACG  IRCAG  IRCGA  IRGAC  IRGCA GACRI  GACIR  GAIRC  GAICR  GARCI  GARIC  GCAIR  GCARI  GCIRA  GCIAR  GCRAI  GCRIA  GIACR  GIARC  GICAR  GICRA  GIRCA  GIRAC  GRAIC  GRACI  GRCAI  GRCIA  GRIAC  GRICA  RACGI  RACIG  RAIGC  RAICG  RAGIC  RAGCI  RCAIG  RCAGI  RCIAG  RCIGA  RCGIA  RCGAI  RIAGC  RIACG  RICAG  RICGA  RIGAC  RIGCA  RGAIC  RGACI  RGCAI  RGCIA  RGIAC  RGICA

RESULTS TABLE FOR FIVE LETTER WORDS:

 NAME Number of different combinations DAVID 60 CRAIG 120

The results table for a five-letter word has proven my prediction.  I

Conclusion

FOUR LETTER WORD:

AABB  ABAB  ABBA  BBAA  BABA  BAAB

FIVE LETTER WORDS:

AAABB  AABBA  ABBAA  ABABA  ABAAB  AABAB   BBAAA  BABAA  BAABA  BAAAB

AAAAB  AAABA  AABAA  ABAAA  BAAAA

RESULTS TABLE:

 NAME Number of different combinations AABB 6 AAABB 10 AAAAB 5

PREDICTION:

I predict that the number of letters in the word divided by the number of A’s divided by the number of B’s will give the number of different combinations.

To test my prediction I will work it out using this rule: for a four letter word like AABB; it has 2 A’s and 2 B’s so: 1x2x3x4 ÷ 1x2 ÷ 1x2=24 ÷ 4=6, and for a five letter word like AAABB; this has 3 A’s and 2 B’s so:  1x2x3x4x5 ÷ 1x2x3 ÷ 1x2=120 ÷ 12=10, and for the five letter word AAAAB; this has 4 A’s and 1 B so: 1x2x3x4x5 ÷ 1x2x3x4 ÷ 1x1=120 ÷ 24=5

This rule explained in algebraic terms is: n! ÷ (A! x B!).  I will now test this algebraic rule for AABB:

4! ÷ (2! x 2!)

24 ÷ (2 x 2)

24 ÷ 4=6 different combinations.

This student written piece of work is one of many that can be found in our GCSE Comparing length of words in newspapers section.

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