# Emma's Dilemma

Extracts from this document...

Introduction

Amy Cotter 11u

Maths Coursework

Investigation:

“Emma’s Dilemma”

Emma’s Dilemma

I am going to investigate the number of different combinations of various groups of letters.

LUCY is a 4 letter word with letters all different. This is the number of combination there can be for LUCY:

LUCY CLUY There are 24 combinations for a 4 letter

LUYC CLYU word that include letters that are all

LCYU CYLU different.

LCUY CYUL

LYUC CULY

LYCU CUYL

ULCY YLUC

ULYC YLCU

UYLC YUCL

UYCL YULC

UCLY YCLU

UCYL YCUL

EMMA is a 4 letter word with 2 letters the same and 2 different. This is the number of combination there can be for EMMA:

EMMA AEMM There are 12 combinations for a 4 letter

EMAM AMME word that includes 2 letters the same and 2

EAMM MMAE different.

MEMA MMEA

MEAM MAEM

AMEM MAME

After investigating to number of combinations there are for the words

Middle

DAEBC

DAECB

DBACE

DBAEC

DBCAE

DBCEA

DBEAC

DBECA

DCABE

DCAEB

DCBAE

DCBEA

DCEAB

DCEBA

DEABC

DEACB

DEBAC

DEBCA

DECAB

DECBA

EABCD

EABDC

EACBD

EACDB

EADBC

EADCB

EBACD

EBADC

EBCAD

EBCDA

EBDAC

EBDCA

ECABD

ECADB

ECBAD

ECBDA

ECDAB

ECBBA

EDABC

EDACB

EDBAC

EDBCA

EDCAB

EDCBA

Now I will do the same again but include 2 letters that are the same in each word, starting with the simplest possible, 2 letters:

AA There is 1 combination for 2 letters that

are both the same.

AAB There are 3 combinations for 3 letters that

ABA include 2 letters the same and 1

BAA different.

AABC BAAC There are 12 combinations for 4 letters

AACB BACA that include 2 letters the same and 2

ABAC BCAA different.

ABCA CAAB

ACAB CABA

ACBA CBAA

There are 60 combinations for 5 letters that include 2 the same and 3 different.

AABCD

AABDC

AACBD

AACDB

AADBC

AADCB

ABACD

ABADC

ABCAD

ABCDA

ABDAC

ABDCA

ACABD

ACADB

ACBAD

ACBDA

ACDAB

ACDBA

ADABC

ADACB

ADBAC

ADBCA

ADCAB

ADCBA

BAACD

BAADC

BACDA

BACAD

BADAC

BADCA

BCAAD

BCADA

BCDAA

BDCAA

BDACA

BDAAC

CAABD

CAADB

CABAD

CABDA

CADAB

CADBA

CBAAD

CBADA

CBDAA

CDAAB

CDABA

CDBAA

DAABC

DAACB

DABAC

DABCA

DACAB

DACBA

DBAAC

DBACA

DBCAA

DCAAB

DCABA

DCBAA

From the previous information, I can form this table which may later help me in finding a formula. In the table below, “s” stands for the number of letters that are the same as each other. There can be more than one number for “s”, for example in this amount of letters:

AABBCC

There are 2 As, 2 Bs and 2 Cs.

Conclusion

AAABCD

AAABDC

AAACBD

AAACDB

AAADBC

AAADCB

AABACD

AABADC

AABCAD

AABCDA

AABDAC

AABDCA

AACABD

AACADB

AACBAD

AACBDA

AACDAB

AACDBA

AADABC

AADACB

AADBAC

AADBCA

AADCAB

AADCBA

ABAACD

ABAADC

ABACAD

ABACDA

ABADAC

ABADCA

ABCAAD

ABCADA

ABCDAA

ABDAAC

ABDACA

ABDCAA

ACAABD

ACAADB

ACABAD

ACABDA

ACADAB

ACADBA

ACBAAD

ACBADA

ACBDAA

ACDAAB

ACDABA

ACDBAA

ADAABC

ADAACB

ADABAC

ADABCA

ADACAB

ADACBA

ADBAAC

ADBACA

ADBCAA

ADCAAB

ADCABA

ADCBAA

BAAACD

BAAADC

BAACAD

BAACDA

BAADAC

BAADCA

BACAAD

BACADA

BACDAA

BADAAC

BADACA

BADCAA

BCAAAD

BCAADA

BCADAA

BCDAAA

BDAAAC

BDAACA

BDACAA

BDCAAA

CAAABD

CAAADB

CAABAD

CAABDA

CAADAB

CAADBA

CABAAD

CABADA

CABDAA

CADAAB

CADABA

CADBAA

CBAAAD

CBAADA

CBADAA

CBDAAA

CDAAAB

CDAABA

CDABAA

CDBAAA

DAAABC

DAAACB

DAABAC

DAABCA

DAACAB

DAACBA

DABAAC

DABACA

DABCAA

DACAAB

DACABA

DACBAA

DBAAAC

DBAACA

DBACAA

DBCAAA

DCAAAB

DCAABA

DCABAA

DCBAAA

From the previous information, I can form this table which may later help me in finding a formula:

Table 3:

Number of Letters (n) | s | Number of combinations (c) |

3 | 3 | 1 |

4 | 3 | 4 |

5 | 3 | 20 |

6 | 3 | 120 |

From tables 1 and 2, I can devise this formula:

n! ÷ s! = c

where s is the number of letters that are the same. s can does not have to be one number, for example in this group of letters:

AABBCC

s would be 2, 2 and 2. When there is more than 1 value for s you multiply them. So using n! ÷ s! = c, I predict that the number ofcombinations for AABBCC is:

n! ÷ s! = c

= 6! ÷ 2! x 2! x 2!

= 720 ÷ 2 x 2 x 2

= 720 ÷ 8

= 90

Here I have tested my prediction that there will be 90 combinations for AABBCC:

AABBCC

AABCBC

AABCCB

AACCBB

AACBCB

AACBBC

ABABCC

ABACBC

ABACCB

ABBACC

ABBCAC

ABBCCA

ABCABC

ABCACB

ABCBAC

ABCBCA

ABCCAB

ABCCBA

ACACBB

ACABCB

ACABBC

ACCABB

ACCBAB

ACCBBA

ACBACB

ACBABC

ACBCAB

ACBCBA

ACBBAC

ACBBCA

BAABCC

BAACBC

BAACCB

BABACC

BABCAC

BABCCA

BACABC

BACACB

BACBAC

BACBCA

BACCAB

BACCBA

BBAACC

BBACAC

BBACCA

BBCAAC

BBCACA

BBCCAA

BCCBAA

BCCABA

BCCAAB

BCBCAA

BCBACA

BCBAAC

BCACBA

BCACAB

BCABCA

BCABAC

BCAACB

BCAABC

CAACBB

CAABCB

CAABBC

CACABB

CACBAB

CACBBA

CABACB

CABABC

CABCAB

CABCBA

CABBAC

CABBCA

CBBCAA

CBBACA

CBBAAC

CBCBAA

CBCABA

CBCAAB

CBABCA

CBABAC

CBACBA

CBACAB

CBAABC

CBAACB

CCAABB

CCABAB

CCABBA

CCBAAB

CCBABA

CCBBAA

This proves my prediction correct, because there are 90 combinations for AABBCC.

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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