• 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
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  • Level: GCSE
  • Subject: Maths
  • Word count: 3184

Emma's Dilemma

Extracts from this document...

Introduction

        GCSE Maths Coursework        

Emma’s Dilemma

Introduction

        Emma and Lucy are playing with arrangements of the letters of their names, on arrangement of Lucy is:

                                LUCY

Another arrangement is:

                                YLCU

I will be investigating all the various arrangements of ‘Lucy’ and all the arrangements possible for ‘Emma’

I will also investigate the various arrangements of words with different lengths and different letters, i.e. 3 lettered words, 4 lettered words, 5 lettered words, etc. I hope to then find a pattern which will then help me take my investigation further. I will split my investigation into three parts to make it more systematic and easier to understand. The first part will consist of me finding the various arrangements of ‘Lucy’ by listing them all. The second part will consist of me finding the various arrangements of ‘Emma’ which I will also do by listing. Finally the third part will consist of me trying to find out how many possible arrangements there are for words with various lengths and different letters, as I mentioned above (3 lettered words…).

Part 1

I will be listing all the various possibilities for ‘Lucy’

Lucy

Whilst listing the different possibilities I will try to be as systematic as I possibly can.

LUCY                ULCY                CLUY                YLCU        

LUYC                ULYC                CLYU                YLUC

LCYU                UCLY                CYLU                YULC                24 POSSIBILITIES

LCUY                UCYL                CYUL                YUCL

LYCU                UYLC                CUYL                YCUL

LYUCUYCL                CULY                YCLU

In the effort of trying to be systematic, whilst I was listing the various arrangements I decided to split it into four parts, each part for a different letter of the four lettered name. This is indicated above.  

Part 2

I will now be listing all the different possibilities for ‘Emma’

Emma

Whilst listing the various possibilities I will once again try to be as systematic as I possibly can.

...read more.

Middle

2 * 3

= 6        NUMBER OF POSSIBILITIES

        I have colour coordinated this calculation so that I can explain it. The 2 is the previous answer I obtained, which can be checked in the table plotted above. The 3 is the number of letters in the word, and the 6 indicates all the various possibilities. As you can see, this answer is correct and can be double-checked in the table above.

        I will also do another example to double-check that my method works, and that the pattern is correct. Therefore I will be using the four lettered word, and the outcome should be 24, as I have worked this out previously using the listing method. My working is shown below:

6 * 4

= 24         NUMBER OF POSSIBILITIES

        This proves that my pattern is correct, and that the method works. This means that I can add on to my table of discoveries, using my method of working. Using the method I will now do a five lettered word, and then a six lettered word. My working will be shown below:

Five Lettered Word

24 * 5

= 120         NUMBER OF POSSIBILITIES

Six Lettered Word

120 * 6

= 720NUMBER OF POSSIBILITIES

        I will now show you a development of my table.

Number Of Letters In The Word

Number Of Possible Outcomes

1

1

2

2

3

6

4

24

5

120

6

720

        Whilst doing my working I also managed to find another pattern. This is shown on the following page.

1        2        3        4        5        6

1        2        6        24        120        720

         1 * 1 * 2

1 * 1

1 * 1 * 2 *3            

                                                      1 * 1 * 2 * 3 * 4  

1 * 1 * 2 * 3 * 4 * 5

                 1 * 1 * 2 * 3 * 4 * 5 * 6

I then showed all my discoveries to my teacher, who then pointed out that what I had worked out is known as a factorial. I was then introduced to the factorial button on the calculator. This would enable me to workout the number of arrangements possible for a word with N number of different letters. I have come up with a conclusion of:

N!

...read more.

Conclusion

4! / 4!

        Therefore I decided to multiply them together, hence giving me the equation mentioned above:

4! / (2! * 2!)

= 6

        The only way to prove whether or not this is correct is by listing all the possible outcomes, and hence I have done so on the following page.

AABB

AABB                BBAA

ABAB                BABA        6 POSSIBILITIES

ABBABAAB

I tried to be as systematic as I possibly could, and the number of outcomes is indicated above.

        According to this my prediction seems to be correct, and hence I have managed to come up with another generalised equation, which can be used to work out the various combinations possible with a word with any length and more than one letter repeated. This is shown below:

AAA…ABBB…B

ST

N!

        (S! * T!)

        An example is ‘AAAABBB’

7! / 4! * 3!

= 35NUMBER OF POSSIBILITIES

        The equation can also be extended if necessary. This is done if there are more than two letters that are repeated, for example ‘AABBCCDD.’ The equation in this case will be:

                    N!

OR  8! / (2! * 2! * 2! * “

         (S! * T! * U! * V!)        =2520

Conclusion

        In conclusion I managed to find three different formulas which are developments of each other. As I worked on each part of the investigation I added a little part to the equation, and by the end of part five I have ended up with the final equation. This can be used to find all the possible combinations of any particular word, no matter how many repeats it has, nor how many letters are repeated. For example I can work out all the possible outcomes for a random word such as ‘MISSISSIPPI’ a shown below. Before I work out the suggested word I will show you the evolution of my equation.

  1. N!
  2. N! / S!
  3. N! / (S! * T!)

        The working used to find how the number of possible outcomes for the word ‘MISSISSIPPI.’

Equation = N! / (S! * T! * U!)

        = 11! / (4! * 4! * 2!)

        = 34650

...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

    would only waste time, and effort. I am confident in my rule, and the maths behind it so I haven't written down all of the arrangements. Justification The reason why this rule occurs, is because you take the number of arrangements for the previous number of letters, and because there is one extra letter, you times

  2. Maths GCSE Coursework: Emma's Dilemma

    So in conclusion, we have 3 starting letters and 2 combinations for each one. So it's 3x2x1=6 combinations All letters in brackets can be rearranged 4 letters: A/B/C/D A (BCD) B (ACD) C (ABD) D (ABC) The same method is required her, except now there is another letter (D).

  1. EMMA'S DILEMMA

    DDAYD 15. ADDYD 6. DDYAD 16. ADDDY 7. DDYDA 17. YDDDA 8. DDDAY 18. YDADD 9. DDDYA 19. YDDAD 10. DYADD 20. YADDD There are 20 arrangements in a 5-lettered word with 3 same letters. Let's use the formula and check if it is correct. 1*2*3*4*5 / 1*2*3 = 20 The formula works.

  2. Emma's Dilemma

    LUCY) and divide it by 2. 24 ? 2 = 12 different arrangements. Why you divide by 2: When finding the number of different arrangements for words that have one letter repeated twice, you halve the number of different arrangements for a word which has the same amount of letters, but none repeated.

  1. Emma's Dilemma

    60 20 5 I think that a name with a name with four-letters the same will have a fourth of the arrangements that the triple letter name has. This would make it a 24th of the arrangement with no repeats (this is the same as 4!).

  2. EMMA's Dilemma Emma and Lucy

    11221212 11121222 11212122 11221221 11122122 11212212 11222112 --------- 15 arrangements 11122212 11212221 11222121 11122221 11221122 11222211 12111222 12122112 12212121 12121212 12112122 12122121 12212211 12211212 12112212 12122211 12221112 --------------20 arrangements 12112221 12211122 12221121 12121122 12211221 12221211 12121221 12212112 12222111 2------- so on ------35 arrangement Total arrangement is 70, it works the

  1. Emma's Dilemma

    However, there were only 4 different arrangements in which they were not: BA1A2A3 A1BA2A3 A2BA1A3 A3BA1A2 BA1A3A2 A1BA3A2 A2BA3A1 A3BA2A1 BA2A1A3 A1A2BA3 A2A1BA3 A3A1BA2 BA2A3A1 A1A2A3B A2A1A3B A3A1A2B BA3A1A2 A1A3BA2 A2A3BA1 A3A2BA1 BA3A2A1 A1A3A2B A2A3A1B A3A2A1B BAAA ABAA AABA AAAB Again, to credit my findings, I investigated a similar word "mooo".

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

    This allows the formula to be used to solve every question which needs to know the number of arrangements. The reason why this wasn't added to the previous equation of "Two letters repeated twice": No. of = No. of letters X ( No.

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