# GCSE Mathematics Coursework - Emma's Dilemma

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Introduction

GCSE Mathematics Coursework - Emma's Dilemma

Method:

First, I tested different arrangements of the name 'EMMA', by systematically rearranging the letters in the name such as:

1-EMMA 4-AEMM 7-MMEA 10-MEAM

2-EAMM 5-AMME 8-MEMA 11-MAEM

3-EMAM 6-AMEM 9-MMAE 12-MAME

I found that there were twelve arrangements for the name 'EMMA'. I then investigated different arrangements of letters in the name 'LUCY', using the same method:

1-LUCY 7-UCYL 13-CYLU 19-YCLU

2-LUYC 8-UCLY 14-CYUL 20-YCUL

3-LCUY 9-UYLC 15-CUYL 21-YUCL

4-LCYU 10-UYCL 16-CULY 22-YULC

5-LYCU 11-ULCY 17-CLYU 23-YLCU

6-LYUC 12-ULYC 18-CLUY 24-YLUC

I found that there were many more arrangements in the name 'LUCY' than in the name 'EMMA'. This is because the name 'EMMA' contains repeated letters and the name 'LUCY' does not. I then investigated how many different arrangements of letters there were in other names of different lengths. The lengths of names that I used were 2, 3, and 4 letter names. I recorded my results in a table.

Middle

10

3628800

I then investigated more names, which had repeated letters, and found that there were fewer arrangements of letters. There also appeared to be a pattern involving factorials in this case, although it was not the same pattern as was found in names with no repeated letters. I recorded my results in a table.

Table showing the number of different arrangements of letters possible in words of different lengths with no repeated letters

Number of letters | Number of different arrangements |

2 | 1 |

3 | 3 |

4 | 12 |

I found that the numbers in these results were in the same pattern as the results for the names with no repeating letters, but the numbers of arrangements of the names with no repeating letters was halved for example:

Number of letters | Arrangements (letter repeated once) | Arrangements (no repeating letters) |

1 | - | 1 |

2 | 1 | 2 |

3 | 3 | 6 |

4 | 12 | 24 |

So, the formula for the number of arrangements for words containing two repeated letters (for example, sArAh) is:

(Number of letters in word)!

2!

I then investigated the number of different arrangements of letters in a word with tree repeated letters, and, once again found a similar pattern. By testing the name ‘PIPPA’, I found this set of results:

Number of letters | Arrangements (letter repeated once) |

1 | - |

2 | - |

3 | 1 |

4 | 4 |

5 | 20 |

Conclusion

Number of letters | Number of sets of repeated letters | Number of repeated letters | Number of possible arrangements |

4 | 2 | 4 | 6 |

5 | 2 | 4 | 30 |

I continued to investigate different formulae which could be used to find the number of arrangements in a word with two sets of repeated letters.

A four-letter word such as XXYY has 2 X's and 2 Y's. Therefore 1x2x3x4 (as there are 4 letters) or (1x2)x(1x2) will give the number of different arrangements when:

Number of letters in word (n) factorial

(number of X's)!x(number of Y's)!

or

n!

(Number of x!)x(number of y!)

I then used this formula to calculate the number of arrangements possible in words of length 6,7 and 8 letters containing two sets of repeated letters. I also tested words which contained different numbers of X's and Y's, for example: XXXYY or XXYYY. These results are also shown in the table below. I also calculated some values for other words with greater numbers of letters.

Number of letters | Number of sets of repeated letters | Number of repeated letters | Number of possible arrangements | |

X | Y | |||

4 | 2 | 2 | 2 | 6 |

5 | 2 | 2 | 2 | 30 |

6 | 2 | 2 | 2 | 180 |

7 | 2 | 2 | 2 | 1260 |

8 | 2 | 2 | 2 | 10080 |

5 | 2 | 3 | 2 | |

6 | 2 | 3 | 2 | |

7 | 2 | 3 | 2 | |

7 | 2 | 4 | 3 |

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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