# To investigate the combination of arrangement of letters in Jeans name and then for her friend Emma.

Extracts from this document...

Introduction

Jean’s DilemmaPAGE 1

AIM: To investigate the combination of arrangement of letters in Jeans name and then for her friend Emma.

Plan: First of all I will systematically write down the different combinations to Jeans name and then for her friend Emma.

Investigation: Before I write the combinations down I will first make a prediction to the number of combinations. I predict there will be 24 combinations to Jeans name.

JEAN:

J E A N J E N A J N E A J N A E J A N E J A E N A J E N A J N E A E J N A E N J A N J E A N E J E J A N E J N A E N A J E N J A E A J N E A N J N J E A N J A E N E J A N E A J N A J E N A E J

There are 24 combinations, my prediction was correct, I will now try the combinations for Jeans friend, Emma, once again I predict there will be 24 combinations.

EMMA:

E M M A E M A M E A M M M A M E M A E M M M E A M M A E M E M A M E A M A M M E A M E M A E M M

Middle

JAMES: I predict there will be 100 combinations to the name James.

J A M E S J A M S E J A S M E J A S E M J A E S M J A E M S J M A E S J M A S E J M S E A J M S A E J M E S A J M E A S J E S A M J E S M A J E M A S J E M S A J E A M S J E A S M J S A E M J S A M E J S M A E J S M E A J S E A M J S E M A

So far there are 24 combinations and I have only done one letter at the beginning, so I have a theory that if you substitute the J for any other letter that will give you another 24 combinations, so that’s would mean there are

PAGE 3

5 x 24 combinations = 120. If this theory is right then that would mean 4 letters would be 4 x the number of combinations of a 3 letter word. And then for 3 it would be 3 x the number of combinations of a 2 letter word, and then again the same for 2, 2 x the number of combinations of 1 letter. Which would mean 5 is, 5 x 4 x 3 x 2 x 1 = 120. And 4 x 3 x 2 x 1 = 24.

Conclusion

There are 20 combinations which is 120 / 3 / 2 / 1 / 1.

So

JAMES is 120 / 1 / 1 / 1 / 1 / 1.

SARAH is 120 / 2 / 1 / 1 / 1.

BILLI is 120 / 2 / 2 / 1.

But…..

OZZZY is 120 / 3 / 2 / 1 / 1 .

OZZZO is 120 / 3 / 2 / 2.

Using this table I saw that when there is 3 letters you divide by 3 then 2, so maybe you actually divide by 1 as well but it makes no difference to the answer. So this would change the table slightly to

JAMES is 120 / 1 / 1 / 1 / 1 / 1.

SARAH is 120 / {2 / 1} / 1 / 1 / 1.

BILLI is 120 / {2 / 1} / {2 / 1} /1.

OZZZY is 120 / {3 / 2 / 1} / 1 / 1 .

OZZZO is 120 / {3 / 2 / 1} / {2 / 1}.

PAGE 5

Using this whole theory you can work out any number of letters with any combination of the same letter, I will draw 2 charts to show this for 10 letters and 7 letters. To show 10x9x8x7x6x5x4x3x2x1 I will write 10! As shorthand. And the same for equations like 3/2/1 or 5/4/3/2/1.

10 letters | |||

number of different letters | Same ones | Calculation | Combinations |

10 | All different | 10!/1/1/1/1/1/1/1/1/1/1 | 3628800 |

9 | 1 1 1 1 1 1 1 1 2 | 10!/2!/1/1/1/1/1/1/1/1/1 | 1814400 |

8 | 1 1 1 1 1 1 2 2 | 10!/2!/2!/1/1/1/1/1/1 | 907200 |

7 | 1 1 1 1 2 2 2 | 10!/2!/2!/2!/1/1/1/1 | 453600 |

6 | 1 1 2 2 2 2 | 10!/2!/2!/2!/2!/1/1 | 226800 |

5 | 2 2 2 2 2 | 10!/2!/2!/2!/2!/2! | 113400 |

4 | 3 3 2 2 | 10!/3!/3!/2!/2! | 25200 |

3 | 3 3 3 1 | 10!/3!/3!/3!/1 | 16800 |

2 | 5 5 | 10!/5!/5! | 252 |

7 letters | |||

number of different letters | Same ones | Calculation | Combinations |

7 | All different | 7! | 5040 |

6 | 1 1 1 1 1 2 | 7!/2! | 2520 |

5 | 1 1 1 2 2 | 7!/2!/2! | 1260 |

4 | 1 2 2 2 | 7!/2!/2!/2! | 630 |

3 | 3 2 2 | 7!/3!/2!/2! | 210 |

2 | 4 3 | 7!/4!/3! | 35 |

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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