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  • Level: GCSE
  • Subject: Maths
  • Word count: 2064

Emma's Dilemma

Extracts from this document...

Introduction

Maths Coursework – Emma’s Dilemma

Emma’s Dilemma

Plan

I am investigating the number of different arrangements there will be in different types of names. Some names I will investigate on will have no identical letters such as LUCY. Some will have a pair of identical letters such as EMMA. Some names will have different quantities of letters such as AMMIE, JOE and ANNE.

Firstly, I’ll produce a method which will help me figure out the different arrangements in the name EMMA and LUCY without using any formulas. Using this method I’ll test all my predictions.

I’ll create formulas for different types of names. Examples are names with no identical letters, a pair or more identical letters and names that have 2 different groups of identical letters

Throughout the investigation I will devise different formulas for working out different types of names. By using one of these formulas I’ll figure out the total number of arrangements of the letters XX……XXYY…….Y.

To start of the investigation I have to find the different arrangements of letters in the name EMMA and LUCY without the use of any types of formulas. Therefore, I have developed my own method.

Method

  1. Write the name in its original format. In this case it will be EMMA or LUCY.
...read more.

Middle

JOE= 3X(2X1)= 6

JOHN= 4X(3X2X1) = 24

By looking at this pattern the arrangements of a 5-letter word will be = 5X(4X3X2X1) = 120

A 6 letter word with no identical letters would equal = 6X(5X4X3X2X1) = 720

These type of sequences are called number FACTORIALS, which are represented by the ! Button on the scientific calculator.

So the formula for calculating the number of different arrangements in a word with no identical letters is:

A=N!

A = total number of arrangements

N = the number of letters in the word.

Testing and predictions

To test if the formula N! is correct I will use it to figure out the number of total arrangements in the name JOE.

Prediction:

There are 3 letters in the name JOE.

So 3! = 3X2X1 = 6

Testing

Starting letter J

Starting letter O

Starting letter E

JOE

JEO

OJE

OEJ

EJO

EOJ

Overall Total: 6

My Prediction is correct!

What if a name had a pair of identical letters

From the name EMMA, I already know a 4 letter word with a pair (2 identical letters) of identical letters can be arranged 12 times.

To make this theory correct I will investigate the number of arrangements there will be in the name ANNE.

Starting letter “A”

Starting letter “N”

Starting letter “E”

ANNE

ANEN

AENN

NANE

NAEN

NEAN

NENA

NNEA

NNAE

EANN

ENAN

ENNA

Total: 3

Total : 6

Total: 3

Overall total: 12

Arrangements in AMMIE’s name

Starting Letter “A”

Starting Letter “M”

Starting Letter “I”

Starting Letter “E”

AMMIE

AMMEI

AMIME

AMIEM

AMEMI

AMEIM

AIMME

AIMEM

AIEMM

AEMMI

AEMIM

AEIMM

MAMIE

MAMEI

MAIME

MAIEM

MAEMI

MAEMA

MMAIE

MMAEI

MMIAE

MMIEA

MMEAI

MMEIA

MIAME

MIAEM

MIMAE

MIMEA

MIEMA

MIEAM

MEAMI

MEAIM

MEMIA

MEMAI

MEIAM

MEIMA

IAMME

IAMEM

IAEMM

IMAME

IMAEM

IMMAE

IMMEA

IMEAM

IMEMA

IEAMM

IEMAM

IEMMA

EMMAI

EMMIA

EMAMI

EMAIM

EMIMA

EMIAM

EAMMI

EAMIM

EAIMM

EIAMM

EIMAM

EIMMA

Total: 12

Total: 24

Total: 12

Total: 12

Overall Total: 12X3+24= 60

...read more.

Conclusion

There are many ambiguous answers I got through my investigation.

Firstly, I tried the formula; A=N!/X! on the name Lucy. Since there wasn’t any identical letters in the name, I substituted X for 0 which gave me the whole answer of 0. This gave me the idea that the A=N!/X! only suitable for words with 2 identical letters.

Secondly, I tried the same formula for the rows; XX……XXYY…….Y, I got the equation of 20!/10!, which gave me the ambiguous answer of 670442572800. This made me realise I was wrong so I created a new formula to help me figure this row out.


...read more.

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