# From Emma's name, a four letter word, we this time only get twelve arrangements. This exactly half of twenty-four,(twenty-four being the total number of arrangements for a four letter word with all the letters differ

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Introduction

## Coursework

Part 1:

Lucy Lyuc Lcyu Lcuy Lycu Luyc

Ulcy Uylc Ucyl Uycl Ulyc Ucly

Cluy Cylu Cuyl Cyul Clyu Culy

Yluc Yclu Yucl Ycul Ylcu Yulc

If a word has: 2 letters it has 2 combinations. E.g. AB, BA

3 letters it has 6 combinations. E.g. ABC, ACB, BAC, BCA, CAB, CBA

The 6 combinations are found by multiplying the 3 letters times the 2 previous combinations. This works because the combination of the 2 letters only has 2 different combinations and because now there is another letter I just times the amount of letters in the new word by the amount of combinations in the previous word to find the amount of combinations in the new word.

So if the number of combinations to a 3 letter word I multiply together 3 and 2 too give the answer it becomes 3x2x1. And to get the amount of combinations of a for a 4 letter word I multiply together 4x3x2x1 to get the number of combinations. And that continues and work, but only when all the letters in the word are different.

Middle

Number of letters | Number of letters repeated | Arrangements | Calculation |

2 | 2 | 1 | 2! divided by 2 |

3 | 2 | 3 | 3! divided by 2 |

4 | 2 | 12 | 4! divided by 2 |

5 | 2 | 60 | 5! divided by 2 |

6 | 2 | 360 | 6! divided by 2 |

7 | 2 | 2520 | 7! divided by 2 |

Part 3:

However, this formula was ineffective when trying to work out the number of arrangements for a word, which has three letters the same in it, and all the rest different.

AAAB AABA

ABAA BAAA

The word produced four arrangements, which is 24 divided 6(twenty-four being the number of arrangements of a four letter word with all the letters different, and six being the total number of arrangements of a three letter word and three is the number of letters repeated.) I worked out that the correct formula for this was…

4!

6

because if all the A’s were different e.g. A(1) A(2) A(3)

A(1) A(3) A(2)

A(2) A(1) A(3)

A(2) A(3) A(1)

A(3) A(2) A(1)

A(3) A(1) A(2)

These would all be different arrangements but seeing as they aren’t they have to be cancel out from 4!

Conclusion

Arrangements:

XXYY | YYXX |

XYXY | YXYX |

XYYX | YXXY |

Total Number of Arrangements: 6

Name: “XXXYYY”

Number of letters: 6

Arrangements:

XXXYYY | XYYXYX | YYYXXX | YXXYXY |

XXYYYX | XYYXXY | YYXXXY | YXXYYX |

XYYYXX | XYXXYY | YXXXYY | YXYYXX |

XXYXYY | XYXYXY | YYXYXX | YXYXYX |

XXYYXY | XYXYYX | YYXXYX | YXYXXY |

Total Number of Arrangements: 20

Name: “XXYYY”

Number of letters: 5

Arrangements:

XXYYY | XYXYY | XYYYX | YYXXY | YYXYX |

XYYYX | XYYXY | YYYXX | YXYYX | YXXYY |

Total Number of Arrangements: 10

Summary of Part 4:

When there is only one letter repeated, the formula obtained was n!/y! (y being the number of repeated letters)

Therefore, for words where more than one letter is repeated, I will try the formula

Examples:

Name: “XXXYYY”

Number of letters: 6

Name: “XXYYY”

Number of letters: 5

This equation appears to be correct.

- We have therefore arrived at the formula

- The n! is there as the total number of letters, as before. We previously decided that where one letter is repeated a certain number of times, the equation would be

- It is therefore logical to assume that where more than one letter is repeated, n! should be divided by the total number of the first repeated letters multiplied by the total number of the second repeated letters. So the over all formula is n!/X! x Y! e.c.t.

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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