• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  • 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.

This student written piece of work is one of many that can be found in our GCSE Emma's Dilemma section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Emma's Dilemma essays

  1. Emma's Dilemma

    and see if any pattern arises. Case Study 1 (2 letters same, all else different) I will start by looking at just one pair of the same letter, with nothing following it: 2 letters: A/A A 1 combination We will now move onto looking at the same pair but with another letter(s)

  2. Emma's Dilemma

    For example, the 20 indicates that there are 20 possible outcomes when there are 3 repeats in a five lettered word (AAABC). The red arrow will represent division, and the blue arrow will represent the equals sign like before. I will do a calculation to try and prove this pattern.

  1. Emma's Dilemma

    / Y! To make sure this is correct I will see if it works by using the formula to check each set of results I have. Double letter, double letter = (4! / 2!) / 2! = 6 Double letter, triple letter = (5!

  2. Emma's Dilemma

    into the number of different arrangements for the letters in the word "mooo". On the whole, there were only really four different arrangements for the letters in the word but for each of these, there were six identical arrangements where the individual O's were arranged in a different way.

  1. Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

    Answer: From the answers I have learned from the previous parts of this coursework, I should be able to produce a general rule to solve any question which follows the same basic patterns of the sequences of letters in the earlier parts of this coursework.

  2. I have been given a problem entitled 'Emma's Dilemma' and I was given the ...

    For AABB I worked out, I did not take into consideration the second set of double letters. The number of permutations (6) by dividing the number of letters factorial (4! or 24) by 4, that is: 6 (i.e. Y) = 24 (i.e.

  1. EMMA'S DILEMMA

    So the formula works. But lets see if it will work on a 5-lettered word with 3 same letters. 5-lettered word with 3 same letters: 1. DADDY 11. DYDDA 2. DADYD 12. DYDAD 3. DAYDD 13. AYDDD 4. DDADY 14. ADYDD 5. DDAYD 15. ADDYD 6. DDYAD 16. ADDDY 7.

  2. Emma's Dilemma

    Test and prediction Prediction: To test out my formulae; A=N!/X! I will test it out on the name OTOJO In this name there are 5 maximum letters and 2 identical letters so: A=5!/3! = (5X4X3X2X1)/(3X2X1) = 120/6 = 20 Test Arrangements in OTOJO's name Starting letter "O" Starting letter "T"

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work