# 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.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month