# Investigate all the possible combinations there are of a person's name, using only the letters in their name.

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Introduction

Emma’s Dilemma

Introduction

This piece of course work was set to investigate all the possible combinations there are of a person’s name, using only the letters in their name.

For example Joe:

The combinations are

Joe

Jeo

Eoj

Ejo

Oej

Oje

Part 1

Lucy

In this section, I will be exploring all the combinations of the name Lucy and if there is any correlation between the number of combinations and the number of letters in her name.

Possible Combinations:

- lucy
- luyc
- lcuy
- lcyu
- lyuc
- lycu
- ulcy
- ulyc
- uylc
- uycl
- ucly
- ucyl
- cluy
- clyu
- culy
- cuyl
- cylu
- cyul
- yluc
- ylcu
- yucl
- yulc
- yclu
- ycul

Method

Firstly you choose one letters to start with, then change the second letter, for the example of Lucy we would have

Lucy

Lucy ← The U stays the same but the last two letters change, this pattern is repeated.

From

Middle

Possible Combinations:

- emma
- emam
- eamm
- meam
- maem
- mame
- mema
- mmea
- mmae
- amme
- amem
- aemm

From this we can see that there are 12 possible combinations of the name Emma. This could mean that to find the number of combinations in a name with repeated letters, you take the number of cominations in that amount of letters without repeated letters and divide it by to.

For example

24/2 = 12

As Lucy had 24 combination and when adding a double letter the number of combinations were 12.

I will investigate into this further in Part 3

Part 3 – Further Research

Rohan

I know from investigation that Rohan has 120 combinations of his name, and has 5 letters in his name.

There for it is 24 x 5.

Conclusion

The formula for a name with repeated letters is

X!/R! where R= the amount a letter is repeated.

Using this, the word ABCBDB would be

6!/3!

If a name has 2 sets of repeated letters then you simply divide it by the amount of repeated letters. So a name like Hannah would be

6!/2!2! as there are 2 A’s and N’s.

This is true of any word. For example ABACABCBACABA would be

14!/6!4!3!

Also from using ! we can easily work out all the combinations of a name without having to write them all out:

Number of letters in the word | Amount of possible permutations |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

6 | 720 |

7 | 5040 |

8 | 40320 |

9 | 362880 |

To test this table we can take 9!

Which is 1x2x 3x4x5x6x7x8x9 which is 362880

Tom Edwards 10CHMY – Emmas Dilemma Page of

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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