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• Level: GCSE
• Subject: Maths
• Word count: 1307

# 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

EBACD

EBCDA

EBDAC

EBDCA

ECABD

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

ABACD

ABCDA

ABDAC

ABDCA

ACABD

ACBDA

ACDAB

ACDBA

BAACD

BACDA

BCDAA

BDCAA

BDACA

BDAAC

CAABD

CABDA

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

span> and 3different:

AAABCD

AAABDC

AAACBD

AAACDB

AABACD

AABCDA

AABDAC

AABDCA

AACABD

AACBDA

AACDAB

AACDBA

ABAACD

ABACDA

ABCDAA

ABDAAC

ABDACA

ABDCAA

ACAABD

ACABDA

ACBDAA

ACDAAB

ACDABA

ACDBAA

BAAACD

BAACDA

BACDAA

BCDAAA

BDAAAC

BDAACA

BDACAA

BDCAAA

CAAABD

CAABDA

CABDAA

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.

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