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  • Level: GCSE
  • Subject: Maths
  • Word count: 1752

Emma's Dilemma.

Extracts from this document...

Introduction

James Clappison.

Emma,s Dilemma.

Introduction

A girl called Emma decides that she is going to re-arrange the letters in her name, for example:

EMAM EAMM EMMA MAME

 She tries to find as many different ways to re-arrange her name as possible. She decides that although she has double letters in her name, none of the arrangements can be identical. However, her friend Lucy has four letters in her name, although this is the same as Emma, Lucy has more ways to re-arrange her name.

I will try to find out why this is so. To do this I will find all the different arrangements of both names and also experiment with some names with different numbers of letters, e.g- 2 letters = Al, 3 letters = Sam, 4 letters = Lucy and 5 letters = Katie.

After this I will have all the data I need to calculate a formula. This will tell me how many arrangements of names can be done depending on the length of the name and if any of the letters are the same. Then I shall choose some names with the same numbers of letters, although these names will have some letters the same in them. I will then work out a formula for these.

Investigation 1

...read more.

Middle

Name

No. of letters

No. of arrangements

Formula

Jackie

6

720

6!

Deborah

7

5,040

7!

Dominque

8

40,320

8!

Kathorine

9

362,880

9!

If my previous calculations have been correct, this formula should work for these names also. I will not show any of the arrangements for these to prove my theory because there are too many to write.

Investigation 2

I am now going to investigate names which have 2 letters in them the same. I will show the different arrangements for certain names, and from this data try to calculate a formula.

I am using these names because they have 2 letters the same.

However, I cannot find a name which has 2 letters, and both of them being the same letter. Because of this I shall use AA. Although this is not a name it should still work as well as a name.

2 letter name ~ AA

AA

3 letter name ~ LEE

LEE EEL ELE

4 letter name ~ EMMA

EMMA EMAM EAMM MAME MEMA MMAE MMEA

AMME AMEM AEMM MAEM MEAM

5 letter name ~ ANNIE

ANNIE NNIAE NENIA IENNA EANNI

ANNEI NNIEA NENAI IENAN EANIN

ANEIN NNEIA NEIAN IEANN EAINN

ANENI NNEAI NEINA INNEA ENNAI

ANIEN NNIEA NEANI INNAE ENNIA

ANINE NNIAE NEAIN INEAN ENAIN

AINNE NIEAN NANIE INENA ENANI

AINEN NIENA NANEI INAEN ENIAN

AIENN NINAE NAIEN INANE ENINA

AENNI NINEA NAINE IENAN EINNA

AENIN NIANE NAENI IEANN EINAN

AEINN NIAEN NAEIN IENNA EIANN

...read more.

Conclusion

Name

No. of letters

No. of letters the same

No. of arrangements

Aaa

3

3

1

Aaab

4

3

4

Aaaab

5

3

20

To back up my prediction I will show all the different arrangements of the letters in the table.

3 letter name ~ AAA

AAA

4 letter name ~ AAAB

AAAB

AABA

ABAA

BAAA

5 letter name ~ AAABC

AAABC BCAAA

AAACB CBAAA

ABCAA BACAA

ACBAA CABAA

AABCA CAABA

AACBA BAACA

ACABA BAAAC

ABACA CAAAB

ABAAC ACAAB

This proves that my prediction was right. The formula which I used to work these out is:

a = x!

3!

I worked out this formula by:

Linking it with the two other formulae which I have calculated.

I realised that there were three times as many arrangements of the name when the arrangement started with the letter which was tripled. This was the same in the other experiment, except that it was doubled.

So, again in this formula a three must be in it, and this must be the factorial of three, 3! So, I came up with a formula of:

a = x!

3!

I proved this correct as above. From my three formulae I can predict that to find the number of arrangements with 4 letters the same the formula is:

a = x!

4!

This is a table which shows all the different formulae for these experiments.

No. of letters the same

Formula

0

x!

2

x!

2!

3

x!

3!

4

x!

4!

5

x!

5!

...read more.

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