# Emma's Dilemma.

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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

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

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

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

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