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In this piece of coursework I will investigate how many times and ways I can arrange Emma's name. I will start by looking at Emma's name. Further then I will look at letters with single, double, triple and etc letters on them

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Introduction

IntroductionEmma’s Dilema

In this piece of coursework I will investigate how many times and ways I can arrange Emma’s name. I will start by looking at Emma’s name. Further then I will look at letters with single, double, triple and etc letters on them, and observe how many different combinations I would come up with.

I will make predictions, put my results in tables and look for patterns and outcomes.

I will also use a calculator to check through my figures to make shure they’re correct.

I am now going to be looking at Emma’s name to find out how many different combinations there are for it.

emma

eamm

mmae

mmea

amme

aemm

meam

maem

mame

mema

amem

emam

Emma has 12 different combinations; note that there are 4 total letters and 3 different.

I am now going to find out how many different combinations there are for Lucy.

lucy

lcyu

lycu

luyc

lyuc

lcuy

ucly

ulcy

ucul

ulyc

uycl

uylc

ylcu

yucl

yluc

yclu

yulc

ycul

cyul

cluy

cylu

culy

clyu

cuyl

Lucy has 24 different possibilities in this arrangement, of 4 letters that are all different.

...read more.

Middle

I am now going to find out how many different combinations there are when 2 letters are the same.

ee

eeb

eebo

bee

eeob

ebe

eboe

eobe

beeo

beeo

oeeb

obee

boee

ebeo

eoeb

The different combinations with 2 letters the same

No of letters

Combinations

Fractorial

2

1

2! /2!

3

3

3! /2!

4

12

4! /2!

5

60

5! /2!

6

360

6! /2!

n

n! /2!

I figured out that the factorial nth for different combinations with two letters the same is n!/2!.

I noticed that the number of combinations with 2 letters the same was half the number of combinations with different letters.

 The reason why I think this is because when the total number of letters that have a double letter in it are being arranged the double letter only counts as if it’s one letter.

...read more.

Conclusion

I am now going to investigate the different combinations with 4 letters the same.

ccccc

ccccd

cdccc

ccdcc

cccdc

dcccc

The different combinations with 4 letters the same

No of letters

Combinations

Fractorial

4

1

4! /4!

5

5

5! /4!

6

30

6! /4!

n

n! /4!

The factorial nth term is n! /4!

 I can state that if there are 8 letters with 5 the same then the number of combinations would be:

8! /5! which is 8*7*6*5*4*3*2*1 / 5*4*3*2*1=336

The different combinations with double, triple and etc letters the same

No of letters

Fractorial

Combinations

ccdd

4!/(2!2!)

=4*3*2*1/(2*1*2*1)=6 combinations

ccddd

5!/(2!3!)

=5*4*3*2*1/(2*1*3*2*1)=10 combinatoins

cccddd

6!/(3!3!)

=6*5*4*3*2*1/(3*2*1*3*2*1)=20 combinatoins

cccdddd

7!/(3!4!)

=7*6*5*4*3*2*1/(3*2*1*4*3*2*1)=35 combinations

n

n!/(y!z!)

The factorial nth term for the number of possibilities with double, triple and etc letters the same is n!/(y!z!).

I noticed that the bottom two  fractorial add up to the top fractorial, e.g: 4!=2!+2!. I also obsereved that the donomenator is how many letters there are the same and the numerator is how many letters there are all together

Also for example if I wanted to find out how many different combinations I could get with 2 double and 3 triple different letters together I would then do:

5!/(2!3!) which is 5*4*3*2*1/ ( 2*1*3*2*1)= 10 combinations.

...read more.

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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