# Emma's Dilemma.

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Introduction

## Emma's Dilemma

Emma's Dilemma

In my investigation I am going to investigate the number of different arrangements of letters in a word.

e.g.

Tim

Is one arrangement

Mit

Is another

First I am going to investigate how many different arrangements in the name LUCY, which has no letters the same.

LUCY

LUYC

LYCU

LYUC

LCYU

LCUY

ULCY

UCLY

UYLC

ULYC

UCYL

There are 4 different letters and there are 24different arrangements.

SAM

SMA

MSA

MAS

ASM

AMS

There are 3 different letters in this name and 6 different arrangements.

JO

OJ

There are 2 different letters in this name and there are 2 different arrangements.

UYCL

YCUL

YUCL

YULC

YLCU

YLUC

CYLU

CYUL

CUYL

CULY

CLUY

## Table of Results

Number of Letters | Number of Different Arrangements |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

6 | 720 |

7 | 5040 |

From the table of results I have found out that a 2 letter word has 2 arrangements, and a 3 letter word has 6.

Taking for example a 3 letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements.

In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4!, which is called 4 factorial which is the same as 4 x 3 x 2.

So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8 letter word.

The formula for this is: n! = a

Where n = the number of letters in the word and

a = the number of different arrangements.

Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, 3 letters repeated and 4 letters repeated.

EMMA

AMME

AMEM

EMAM

AEMM

EAMM

MMEA

MMAE

MEMA

MAME

MEAM

MAEM

Middle

3

6

3

1

0

0

4

24

12

4

1

0

5

120

60

20

5

1

6

720

360

120

30

6

7

5040

2520

840

210

42

I have worked out that if you do say 5! = 120, to find out how many different arrangement in a 3 letter word it would be 5! Divided by 6 = 20, so, a 6 letter word with 4 letters repeated would be 6! Divided by 24 = 30, as you can see in the "No letters repeated" column these are the numbers we are dividing by:

2 letters the same = n!

(2x1 = 2)

3 letters the same = n!

( 3x2x1 = 6)

4 letters the same = n!

(4x3x2x1 = 24)

5 letters the same = n!

(5x4x3x2x1 = 120)

6 letters the same = n!

(6x5x4x3x2x1 = 72)

From this I have worked out the formula to fine out the number of different arrangements:

n! = the number of letters in the word

p! = the number of letters the same

Now I am going to investigate the number of different arrangement for words with 2 or more letters the same like, aabb, aaabb, or bbbaaa.

This is a 4 letter word with 2 letters the same, there are 6 different arrangements:

xxyy

I am going to use the letters x and y (any letter)

xxyy xyxy yxxy

xyyx yxyx yyxx

This is a 5 letter word

xxxyy

xxxyy xxyxy xxyxx xyxyx xyxxy

xyyxx yyxxx yxxxy yxyxx yxxyx

There are 10 different arrangements

In the above example there are 3 x's and 2 y's

As each letter has its own number of arrangements i.e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again.

As before, the original formula:

A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's)

Conclusion

· xyyzz

· xyyyz

…And so on…

Here are some arrangements for:

· xyyzz

There are 30 arrangements.

The formula can be changed slightly to

(x + y + z) (x + y + z - 1) (x + y + z - 2) (x + y + z - 3) (x + y + z - 4)

(x) (y) (y - 1) (z) (z - 1)

because it has 5 letters, 3 of which are different.

(1 + 2 + 2) (1 + 2 + 2 - 1) (1 + 2 + 2 - 2) (1 + 2 + 2 - 3) (1 + 2 + 2 - 4)

(1) (2) (2 - 1) (2) (2 - 1)

= 5 x 4 x 3 x 2 x 1

1 x 2 x 1 x 2 x 1

= 5!

1! x 2! x 2!

= 120

4

= 30

This proves that my formula works, and in future the 1! Does not need to be included because in names like THANCANAMOOTOO in which there are 3 letters that are not repeated (H, C, M) we do not need to write

1! x 1! x 1! Because 1! Is equal to simply 1 and this does not have any effect on the formula if we exclude it.

The formula can be simplified to

x!

y! x z! x a! …and so on, depending on the number of letters which are different.

x is the number of letters

y is the number of letters repeated, as are z and a.

This formula can go on forever depending on the number of letters repeated twice or more.

So, for example if in a name like THANCANAMOOTOO we would use the formula to work it out like this:

14!

2! x 3! x 2! x 4!

= 14!

2 x 6 x 2 x 24

= 151,351,200

· It is 14! Because that is the number of letters in the name.

· It is divided by (2! x 3! x 2! x 4!) because the letters T and N are both repeated twice, the letter A is repeated 3 times, and the letter O is repeated 4 times, and hence we multiply the factorials of them by each other.

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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