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Introduction

Emma’s Dilemma

Part One

Investigate the number of different arrangements of the letters of Lucy’s name.

Below is a list of the different combinations of Lucy’s name:

Lucy
Luyc
Lcuy
Lcyu
Lyuc
Lycu

Ucyl
Ucly
Ulcy

Ulyc
Uycl
Uylc

Cylu
Cyul
Culy
Cuyl
Clyu
Cluy

Ycul
Yclu
Yulc
Yucl
Ylcu
Yluc

There are 24 different arrangements of the letters of Lucy’s name. This is because there are 6 possible combinations for each letter starting of the arrangement i.e. there are six combinations of the arrangement starting with Y. There are 4 different letters in Lucy’s name: 4 multiplied by 6 gives us 24.

Part 2

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

Below is a list of the different combinations of Emma’s name:

Emma

Eamm

Emam

Meam

Maem

Mmea

Mmae

Mema

Mame

Amme

Aemm

Amem

Emma’s name produces a bit of a problem. Even though there are the same amounts of letters as there are in Lucy’s name there are only a total of 12 different combinations. This is because Emma’s name contains two letters which are the same. If we take one of the m’s from Emma’s name and replace it with an X we will get 24 different combinations of the letters of Emma’s name:

Exma (Emma)

Eaxm (Eamm)

Exam (Emam)

Emxa (Emma same as Exma)

Eamx (Eamm same as Eaxm)

Emax

Middle

REI

Letters

LUCY

Combinations

LUCY
LUYC
LCUY
LCYU
LYUC
LYCU

UCYL
UCLY
ULCY

ULYC
UYCL
UYLC

CYLU
CYUL
CULY
CUYL
CLYU
CLUY

YCUL
YCLU
YULC
YUCL
YLCU
YLUC

By doing this I found out that a pattern emerged. For 1 letter there was only 1 combination. For 2 letters there were 2 combinations. For 3 letters there were 6 combinations. For 4 letters there were 24 combinations. To find out the number of different arrangements of various groups of letters we just multiple the amount of letters by the integers before it. This is factorial. Factorial is the product of an integer and all integers before it. For example:

Let’s take the word ‘end’ for example. It has 3 letters in the word and so 3! should give us the total amount of different arrangements the word has.

3! = 3*2*1

3! = 6

We can prove this by writing out all the combinations for the word ‘end.’

End

Edn

Den

Dne

Ned

Nde

There are 6 combinations for the word ‘end.’ We may wonder why we have to multiply the integers before three, i.e. 2 and 1 to get the different combinations. If we take ‘E’ as out starting letter, there are three possible places that we can place ‘E’ in the three letter word.

_ _ _

If we place ‘E’ here, there are 2 possible places to place the next letter ‘N.’

_ _ _

If we place the letter ‘N’ here, then there is only one possible where we can place the next letter, which is ‘D.

Conclusion

Letter

I

Combinations

1 combination: I

Letters

FF

Combinations

1 combination: FF

Letters

HJH

Combinations

3 combinations:

HJH

JHH

HHJ

Letters

EMMA

Combinations

12 combinations:

EMMA

EAMM

EMAM

MEAM

MAEM

MMEA

MMAE

MEMA

MAME

AMME

AEMM

AMEM

From this I saw a pattern emerging. As

If we take the word ‘See’ for example. If we do 3! we get 6. However the word ‘see’ has only got 3 different combinations:

See

Ees

Ese

As the word has two E’s repeated we divide 3! first by 2!. This eliminates any combinations which are the same.

This process can be used for words of different letters. Below are a few more examples of this:

1. Letters

TT

Combinations

There is only one different arrangement for the letters ‘TT.’

Why?

There are 2 letters in total so we use 2!. The first T has 2 possibilities where it can be placed, and the second T has only one possibility to where it can be placed. If we multiply the total possibilities together we get 2. Now we must divide the 2! to eliminate all possibilities this will be the same. The final

2

1. Letters

Aab

Combinations

There are 3 different arrangements for the letters ‘aammyy.’

Why?

There are 3 letters in total so we use 3!. If A is the first letter, it has 2 possibilities to where it can be placed. It only has two possibilities because it cannot be placed next to the other A. i.e.

Aab

Baa

Aba

1. Letters

Aabb

Combinations

There are 6 different arrangements for the letters ‘aabb.’

Why?

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

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