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# Emma's Dilemma

Extracts from this document...

Introduction

James Clark 11LG Maths Coursework

14th January 2002

## Emma’s Dilemma

The Problem:

Emma is playing with arrangements of the letters of her name.

One arrangement is

EMMA

A different arrangement is

MEAM

Another arrangement is

AEMM

EMMA AMEM

## EMAM AMME

EAMM

MEMA

MMEA

MMAE

MAME

MEAM

MAEM

AEMM

I have found out that when the letter “E” is at the front of the name Emma, then there are three different combinations. The same rule applies when the letter “A” is at the front of the name Emma. There is an exception though, when the letter “M” is at the front of the name Emma then there are six different combinations.

Overall there are 12 different combinations for the name Emma.

Middle

UCYL YUCL

ULYC YCUL

From this second experiment I have found out that when the letter “L” is at the front of the name Lucy then there are six different combinations. The same rule applies to the letters “U”, “C” and “Y” in the name Lucy.

Overall there are 24 different combinations for the name Lucy

From the first two experiments I have noticed that the name Emma has half the number of combinations when compared with the name Lucy. This maybe because of the repeated letter “M” in the name Emma. To find out whether or not this is true I will investigate a 5-letter word with a repeated letter in it and a 5-letter word with all different letters in it. I will then compare my results for these experiments.

For the third pat of my Investigation I will look at other names of varying lengths and letters. The first name I will investigate will be Penny.

Conclusion

BRNIA BIARN BNRIA RBINA RANIB RIBAN ABRIN ARBNI AINBR ANRBI

BRNAI BAIRN BNRAI RBNIA RANBI RNIAB ABRNI ARNBI AINRB ANIRB

BIRNA BAINR BNARI RBNAI RAIBN RNIBA ABNIR ARNIB AIRNB ANIBR

IBRAN IRANB INRAB NRABI NIBAR

IBRNA IRABN INRBA NRBIA RIABN

IBNAR IABRN INABR NRBAI RIANB

IBNRA IARBN INARB NRIBA NBARI

IBARN IARNB NBRIA NRIAB NABIR

IBANR IABNR NBRAI NIRAB NABRI

IRBAN IANBR NBIAR NIRBA NARBI

IRBNA IANRB NBIRA NIARB NARIB

IRNAB INBAR NBAIR NIABR NAIRB

IRNBA INBRA NRAIB NIBRA NAIBR

From this experiment I have found out that there are 24 different combinations for each letter at the beginning of the name Brian.

Overall there are 120 different arrangements for the name Brian.

I will now investigate the different arrangements with smaller names with repeated letters and without. Because I have noticed that this fact has an effect on the outcome.

I have decided to look at the different arrangements for the name Tom.

TOM

TMO

MOT

MTO

OMT

OTM

From this experiment I have noticed that there are two different combinations for each letter at the start of the name Tom.

Overall there are six different arrangements for the name Tom.

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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1. ## Emma's Dilemma

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2. ## GCSE maths coursework: Emma's dilemma

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1. ## Emma's Dilemma

Whereas 'Lucy' which is also a four lettered word had a different number of outcomes which was 24. This means that when there are repeated letters the number of possible outcomes varies. This led me to take my investigation a step further, and hence I will now have a fourth part to my investigation.

2. ## Emma's Dilemma

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1. ## Emma's Dilemma

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