# Emma's Dilemma

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Introduction

Danny Soopen –“Emma’s Dilemma” - Page of

Danny SoopenGCSE Maths Project – “Emma’s Dilemma”

I have been told to investigate the number of different arrangements of the letters of 'EMMA'. I was then told to investigate the number of different arrangements of the letters of 'LUCY'. After these, I must investigate into different arrangements of letters and find a rule to find the number of different arrangements of letters of any name.

Part 1:

Ways of arranging EMMA’s name:

EMMA | MEMA | MMAE | AMME |

EMAM | MEAM | MAEM | AMEM |

EAMM | MMEA | MAME | AEMM |

For each new beginning letter, there are 3 different possible combinations.

There are a total of 12 possible different combinations.

Part 2:

Ways of arranging LUCY’s name:

LUCY | ULCY | CULY | YUCL |

LUYC | ULYC | CUYL | YULC |

LCUY | UCLY | CLUY | YCUL |

LCYU | UCYL | CLYU | YCLU |

LYUC | UYLC | CYLU | YLUC |

LYCU | UYCL | CYUL | YLCU |

For each new beginning letter, there are 6 different possible combinations.

There are a total of 24 possible different combinations.

This combination consists of all the letters being different. I will also try and find a formula for this arrangement, and number of letters. Once I have discovered these formulae I am going to investigate, other combinations of letters and different amounts of letters. I will then try to discover a link between the formulae to enable me to find a formula for the general case.

There are 24 different possibilities in this arrangement of 4 letters all different. I have noticed that with Lucy beginning each different letter. For example there are 6 arrangements with LUCY beginning with L, and 6 beginning with u and so on 6*4(the amount of letters) gives 24.

Middle

1 1 1 1

We can say that tn = n!

For example:

For FRANK’s name, tn = 5!

tn = 5 x 4 x 3 x 2 x 1

= 120

The formula for the number of combinations for a word where all the letters are different can be obtained, (as with the LUCY example on page 2) by simply writing out all of the permutations. As no letters are repeated, it is a straightforward formula.

There are a total of four letters to be placed into four different positions each time. When the first letter is placed, there are four different places to put it in. When the second letter is being placed, one position is already being taken up by the first letter, and so there are only three places where it can be positioned. When the third letter is being placed, there are only two places in which it can be placed, and the final letter must fill the final space.

In the formula, we start with the number four as there are four letters in the word, and at each stage of the arrangement process, there is one less available place in which to put the letters, so the multiplying number decreases at each stage

Therefore, where there are four letters to be placed, we can summarise it as:

tn = 4 x 3 x 2 x 1

or tn = 4!

(b). Investigating the number of different arrangements of letters in names with varying numbers of letters, where one letter is repeated once.

Conclusion

where n = the number of letters in the word,

and R = the number of times the letter occurs.

Part 4:

Investigating the number of different arrangements of letters where more than one letter occurs more than once.

For example: words such as XXYY, XXYYXX, and XXXYYY

The above examples all have more than one letter that repeats more than one time.

Investigation:

Name: “XXYY”

Number of letters: 4

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

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

Where n = the total number of letters

And R1 = the number of times the first letter occurs

And R2 = the number of times the second letter occurs

Examples:

Name: “XXXYYY”

Number of letters: 6

This formula does not work. I think that the formula could perhaps be

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.

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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