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• Level: GCSE
• Subject: Maths
• Word count: 1099

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

Extracts from this document...

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.

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.

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.

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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