• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Emma's Dilemma.

Extracts from this document...

Introduction

Maths Coursework

Emma’s Dilemma

Investigate the number of different arrangements of the letters of Emma’s name

Introduction:

For this piece of coursework I am going to investigate the number of

arrangements in a word such as: BIT. For the word BIT BIT is an arrangement and TIB is another. After that I will look at words with different numbers of letters to see if there’s a pattern emerging. I will use this pattern to work out some algebraic equations for the number of arrangements in a word such as a 20 letter word. After that I will investigate words with a double letter like : EMMA, to see if there’s pattern for that. I will work out the equation for that and then I will extend the investigation even further.

Part 1: Words with no doubles                                  

2 letter word

image00.png

                I have found there are two combinations for a 2 letter word.

The  equation is 1x2 because there are 2 combinations of letters that can be assigned to the 1st letter and then once you’ve assigned a letter to the 1st letter there’s only 1 letter that can be assigned as the 2nd letter.

1x2=2 combinations

3 Letter word

I have found there are 6 combinations for a 3 letter word. The equation is 1x2x3 because there are 3 combinations of letters that can be assigned as the 1st

...read more.

Middle

IBST= number of arrangements for the 4th letter once the 1st, 2nd and 3rd letters have been assigned.

1x2x3x4.

I have done some research by looking in an A-level text book called A Concise Course In A-Level Statistics written by J. Crawshaw and J. Chambers. In this book I found an expression called Factorial. To get the number of combinations for a word you have to get the product of all the whole numbers from 1 up to that number or in my example up to the number of letters there are in the word. This should give you the amount of combinations in a word with no doubles. Factorial can be represented as !.

For example:

4!= 1x2x3x4

C=n!    

Where C= Total number of combinations.

n= number of letters in word. != Factorial

I predict that a 20 letter word with no doubles will be:

20!=2.432 x 10^18(in standard form).

part 2: Words with 1 double

2 letter word

        I have found there is only one combination for a 2 letter word.

The only calculation that I think so far is the total number of arrangements is

1x2=2. 2÷2=1

3 letter word

MMS        I have found there are 3 combination for a 3 letter word.

MSM  1x2x3= 6. 6÷2=3

SMM

4 letter word

 I

...read more.

Conclusion

Part 5: words with more than 1 set of doubles:

C=n! ÷m! n= number of letters in word, m= how many times the letter is repeated. E.g a double would be 2.

4 letter word:

6 combinations

MMNN

MNMN

MNNM

NNMM

NMNM

NMMN

6 letter word:

MMNNAA

MMAANN

MMANAN

MMNANA

                        12 combinations

AANNMM

AAMMNN

AAMNMN

AANMNM

NNMMAA

NNAAMM

NNAMAM

NNMAMA

Final equation:

n! ÷m!( the equation to find the total number of combinations of a word with one set of repeated letters) can be evolved into a more general equation.

I have noticed that for the following equation above

n! is equal to the number of different letters add the amount of each letter, and then factorialed.

Then the m! part is equal to the amount of different letters times the amount of each different letter and then factorialed.

n! x m!

(n+m)! ÷n! x m!

for example a 4 letter word with 1 double:

AABC

Lets split the equation into 2 halves. The first halve being (n+m)!.

In this part n would = AA=2.

And the m part would in fact be =BC=2.

So you would do:

2+2=4

and then find the factorial of 4

4!=24.

Then to the 2nd part which is the answer that you got from the 1st part ÷ n! x m!.

In this case n would be 2 and m would be made into 2 m’s which will all be 1. for example the equation would be

(n+m)! ÷ n! x m! x m!.

2! x 1! x 1! =2

Then we combine the two halves together.

24÷2= 12 combinations.

AABC

AACB

ACAB

ABAC

ABCA

ACBA

CBAA

CABA

CAAB

BAAC

BCAA

BACA

...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

    Therefore, I have proven that this formula works. Now that we know this works, we can predict what the next results will be: Number of letters All different (ABCD) 1 pair of same letter (AA) 2 pairs of same letters (AABB)

  2. Maths GCSE Coursework: Emma's Dilemma

    (five factorial). We then divide that by two to find the next column, and divide that by two again, to find the last column. Case Study 2 (1 letter repeated many times) Now I am going to look at cases where 1 letter is repeated many times.

  1. 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.

  2. 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!

  1. Emma's Dilemma

    Hence, the "something" is "divided by" and the formula for the number of different arrangements for the letters in a name is equal to the n! divided by a!. Number of letters Number of repetitions Number of different arrangements where the repetitions were distinguished Number of different arrangements where the

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

    Question Four: A number of X's and a number of Y's are written in a row such as XX.........XXYY.........Y Investigate the number of different arrangements of the letters. Answer: In this final part of this coursework, I will be investigating how the relation ship between X's and Y's affect my rule, and to see if I can find easier ones.

  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