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• Level: GCSE
• Subject: Maths
• Word count: 1710

# Emma&amp;#146;s Dilemma

Extracts from this document...

Introduction

Luke Plant        10S8                           Emma’s Dilemma                             11/06/2002

In my investigation I am going to investigate the number of different arrangements of letters in a word any try to find any formulae and patterns.

An example of this would be: -

Tim

And another arrangement is: -

Mit

First I am going to investigate how many different arrangements in words, which have

no letters the same.

DATA

1a)  There are 2 different letters in this name and 2 different arrangements.

JO                OJ

1b)  There are 3 different letters in this name and 6 different arrangements

SAM                SMA                MSA                MAS                ASM                AMS

1c)  There are 4 different letters and there are 24 different arrangements

LUCY                LUYC                LYCU                LYUC                LCYU                LCUY

ULCY                ULYC                UCYL             UCLY                UYLC             UYCL

CULY                CUYL                CYUL                CYLU                CLUY                CLYU

YLUC                YLCU                YUCL                YULC                YCUL                YCLU

.

Table of Results

Number of Letters        Number of Differentnts

2                                2

3                                6

4                                  24

From the table of results I have found out that a 2-letter word has 2 arrangements, and a 3-letter word has 6.  The pattern I see emerging is that of  N! (N being No. of letters)

This is true for all of the words in this table.

2! =      2x1     = 2

3! =    3x2x1   = 6

4! =  4x3x2x1  = 24

PREDICTION

So, by using factorial (!) I can predict that there will be 120 different arrangements for a 5-letter word.

1d)  PROOF

J A M I E *

JAMIE                JAMEI                JAEIM                JAEMI                JAIME                JAIEM

JMAIE                JMAEI                JMEIA                JMEAI                JMIEA                JMIAE

JIMAE                JIMEA                JIEMA                JIEAM                JIAEM                JIAME

JEAIM                JEAMI                JEMAI                JEMIA                JEIMA                JEIAM

So there are 5 starting letters and so therefore 5 x 24* = 120 arrangements.

Another formula for this is: n!

Middle

5x4x3x2x1 = 120 arrangements

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

DATA

2a)  2 letter word, 2 letters repeated, 1 arrangement

AA

2b)  3 letter word, 2 letters repeated, 3 different arrangements

MUM                MMU                UMM

2c)  4-letter word, 2 letters repeated, 12 different arrangements.

EMMA        AMME        AMEM        EMAM        AEMM        EAMM

MMEA        MMAE        MEMA        MAME        MEAM        MAEM

Table Of Results

No. of letters     Number of arrangements

2                        1

3                        3

4                          12

From these results I have found that a 2-letter word has only 1 arrangement.

I have also found that a 3-letter word has 3 arrangements.

The pattern I see emerging is that of  N! / 2.

This works for all of the results on this table

2! / 2  =    (2x1) / 2        =  1

3! / 2  =   (3x2x1) / 2     = 3

4! / 2  =  (4x3x2x1) / 2   = 12

PREDICTION

So by using the formula N! / 2 I predict that there will be 60 arrangements for the 5-letter word.

2d) PROOF

G  E  M  M  A*

GEMMA        GAMME        GAMEM        GEMAM        GAEMM        GEAMM

GMMEA        GMMAE        GMEMA        GMAME        GMEAM        GMAEM

So with the 3 starting letters which are not an 'M' there is 12 arrangements.

Therefore 3 x 12 = 36 arrangements.

M  M  E  A  G*

So with 'M' as the starting letter there are 24 arrangements.

Therefore 36 + 24 = 60 arrangements.

Another Formula for this would be

N! / 2 = a                O R           N! / X! = a

Where N = the number of letters in the word,

a = number of arrangements and

x = number of letters the same

This is why the formula is N!

## There are 5 spaces, which the first letter may be put in as shown above.

Conclusion

5! / (3!2!)  =  5x4x3x2x1 /  ((2x1)x(3x2x1)) = 120 / 12 = 10 arrangements

GENERAL RULES

The 1st general rule I have found to be this

n! = the number of letters in the word

x! = the number of letters the same

or in the case of having more than one group of identical letters the 2nd general rule is:

n! = the number of letters in the word

x!y! = the number of repeated letters the same

PROOF FOR GENERAL RULE

General rule 1 has been proved before in previous proof sections ( 1d, 2d and 3d )

Using data 5a and 5b however can prove general rule 2.

The formula for data 5a is:

N! / X! = a

This substitutes into:

4! / 2!2!  =    (4x3x2x1) / (2x1) x (2x1) =  24 / 2x2 = 24 / 4 = 6 arrangements

Also the formula for data 5b is:

N! / X! = a

This substitutes into:

5! / 3!2! = (5x4x3x2x1) / (3x2x1) x (2x1) = 120 / 6 x 2 = 120 / 12 = 10 arrangements

I conclude by presenting the general rules having been proved:

n! = the number of letters in the word

x! = the number of letters the same

n! = the number of letters in the word

x!y! = the number of repeated letters the same

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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