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Emma's Dilemma.

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Introduction

Maths Coursework 2002

Emma’s Dilemma

Rory O’Connell

The Aim of this investigation is to see how many combinations of letters there are in names and other letter combinations. For example the name Emma has the following combinations:

EMMA

EAMM

EMAM

AMEM

AMME

AEMM

MEAM

MAEM

MEMA

MAME

MMEA

MMAE

In the word EMMA there are 12 possible combinations

These are the possible combinations for the word LUCY:

LUCY         UYLC

LUYC         UYCL

LCUY         UCYL

LCYU         UCLY

LYUC

LYCU

CLYU

CLYU

CULY

CUYL

CYUL

CYLU

YLUC

YLCU

YCUL

YCLU

YULC

YUCL

ULCY

ULYC

In the word LUCY there are 24 possible combinations. Twice the amount of  arrangements in the word EMMA, despite having the same amount of letters.

I then looked at the number of combinations of letters there were in names of varying length:

JO

SAM

FRED

DARYL

GERALD

...read more.

Middle

GERALD

720

I have found an equation, which will tell you the number of letter combinations in each word (except EMMA), it is based on the following idea:

In the word FRED for example, there are four letters. When rearranging the letters, there are four possibilities for where the first letter could be placed, and for each of those four possibilities, there are then another three possibilities for where the remaining three letters can go. This means in total so far there are 12 (4x3) possibilities so far.

For each of these 12 possibilities there are then another two possibilities for where the last two letters can go, that means we now have 24 possible combinations (4x3x2). Finally for each of these 24 possible combinations there is only one place for the last letter to go (4x3x2x1 in total). This idea is represented by a exclamation mark called a factorial.

The equation which represents this idea is:      n!

...read more.

Conclusion

XXZZ         ZZXX       ZXZX

XZXZ        ZXXZ        XZZX

 It has six different combinations.

Also look at the word: XXXYY 

XXXYY   XXYXY  

XXYXX   XYXYX  

XYXXY   XYYXX  

YYXXX   YXXXY  

YXYXX   YXXYX

It has eight different combinations

In my last equation I divided the number of letters by the number of permutations of the repeated letter, I believe if there is more than one group of repeated letter you merely have to multiply the number of permutation for each:

N! / (P! x Q!) = A

Where

N = Number of Letters

P and Q = different groups of repeated letters

A = Amount of combinations

If then you had a row of letters that were in unspecified amounts such as:

XX …… XXYY …… Y

Then you would simply apply the above formula, so :

N! / (X! x Q!) = A

Where

N = Number of Letters

X and Y = different groups of letters

A = Amount of combinations

...read more.

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