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  • Level: GCSE
  • Subject: Maths
  • Word count: 1993

Emma's Dilemma.

Extracts from this document...

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

...read more.

Middle

(p)

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)

...read more.

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.

...read more.

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