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  • Level: GCSE
  • Subject: Maths
  • Word count: 1206

Emma's Dilemma

Extracts from this document...

Introduction

Richard Hanton        Mathematics – Coursework        Page  of

        08/05/07

Emma’s Dilemma

  1. Investigate the number of different permutations of the letters of the name Emma.

I am trying to find the maximum number of possible permutations of the name EMMA. This name has four letters but only three variable letters E, M and A.

Permutations:

EMMA        MMAE        AEMM

EMAM        MMEA        AMEM

EAMM        MAME        AMME

                MAEM

                MEMA

                MEAM

This shows us that there are twelve possible permutations of the letters of the name EMMA.

Emma has a friend called Lucy.

  1. Investigate the number of different permutations of the letters of the name Lucy.

...read more.

Middle

ABDCE        BADCE        CADBE        DACBE        EACBD

ABDEC        BADEC        CADEB        DACEB        EACDB

ABECD        BAECD        CAEBD        DAEBC        EADBC

ABEDC        BAEDC        CAEDB        DAECB        EADCB

ACBDE        BCADE        CBADE        DBACE        EBACD

ACBED        BCAED        CBAED        DBAEC        EBADC

ACDBE        BCDAE        CBDAE        DBCAE        EBCAD

ACDEB        BCDEA        CBDEA        DBCEA        EBCDA

ACEBD        BCEAD        CBEAD        DBEAC        EBDAC

ACEDB        BCEDA        CBEDA        DBECA        EBDCA

ADBCE        BDACE        CDABE        DCABE        ECABD

ADBEC        BDAEC        CDAEB        DCAEB        ECADB

ADCBE        BDCAE        CDBAE        DCBAE        ECBAD

ADCEB        BDCEA        CDBEA        DCBEA        ECBDA

ADEBC        BDEAC        CDEAB        DCEAB        ECDAB

ADECB        BDECA        CDEBA        DCEBA        ECDBA

AEBCD        BEACD        CEABD        DEABC        EDABC

AEBDC        BEADC        CEADB        DEACB        EDACB

AECBD        BECAD        CEBAD        DEBAC        EDBAC

AECDB        BECDA        CEBDA        DEBCA        EDBCA

AEDBC        BEDAC        CEDAB        DECAB        EDCAB

AEDCB        BEDCA        CEDBA        DECBA        EDCBA

There are 120 permutations with five different letters

My prediction was correct.

...read more.

Conclusion

1 is the number of occurrences of the first letter, L2 is the number of occurrences of the second letter, L3 is the number of occurrences of the third letter and Ln is the number of occurrences of the last letter.

Therefore I predict that the number of permutations for AABBC will be:

                        P =  _          5!          _          =  120  =  30

                                    2! x 2! x 1!                4

Where T = 5

        L1 = 2

        L2 = 2

        Ln = 1

With two pairs of repeating letters and one different letter.

AABBC        BBAAC        CAABB

AABCB        BBACA        CABAB

AACBB        BBCAA        CABBA

ABABC        BAABC        CBAAB

ABACB        BAACB        CBABA

ABBAC        BABAC        CBBAA

ABBCA        BABCA

ABCAB        BACAB

ABCBA        BACBA

ACABB        BCAAB

ACBAB        BCABA

ACBBA        BCBAA

There are thirty permutations with two pairs of repeating letters and one different letter.

My prediction was correct.

...read more.

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