# Emma's Dilemma

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Introduction

Mathematics GCSE coursework: Emma's Dilemma

In my investigation I am going to investigate the number of possible arrangements of the letters in peoples names.

I start with the least possible amount of letters, which is one:

J

This clearly only has one arrangement.

I will now go on to investigate the number of arrangements of letters in names going up in order of the number of letters that they contain:

AL

LA (2 arrangements)

SAM

SMA

ASM

AMS

MSA

MAS (6 arrangements)

LUCY

LCUY

LCYU

LUYC

LYCU

LYUC

ULYC

ULCY

UCLY

UCYL

UYCL

UYLC

CLUY

CLYU

CUYL

CULY

CYLU

CUUL

YLUC

YLCU

YULC

YUCL

YCUL

YCLU (24 arrangements)

I can see that if the name has no letters that are repeated I can work out the number of different possibilities more easily. I have worked out that when I have exhausted the different number of possible combinations starting with the same letter I can times that by the number of letters in the word to work out the number of formulas. For example, when doing the name Lucy I can cut down the time it would take to work out the name.

LUCY

LCUY

LCYU

LUYC

LYCU

LYUC

SIMON

SIMNO

SIOMN

SIONM

SINOM

SINMO

SMION

SMOIN

SMINO

SMONI

SMNOI

SMNIO

SOIMN

SOINM

SOMIN

SOMNI

SONIM

SONMI

SNIMO

SNIOM

SNMIO

SNMOI

SNOIM

SNOMI

(There are 24 combinations, so taking into account that this number would be possible starting with each letter there will be 24*5 combinations which equals 120 combinations)

Here is a table containing all of the results that I have recorded. Hopefully when I have rounded them all together I might be able to work out a recurring pattern.

Number of letters (N) | Number of combinations |

1 | 1 |

2 | 2 |

3 | 6 |

4 | 24 |

5 | 120 |

Middle

EMMMA

EAMMM

MEAMM

MEMAM

MEMMA

MMEMA

MMMEA

MMEAM

MMMAE

MMAME

MMAEM

MAEMM

MAMEM

MAMME

AEMMM

AMEMM

AMMEM

AMMME (3 letters the same, 20 combinations)

On the following page is a table containing the results for the words that I have just used. Hopefully, by studying these, a pattern will become clearer and I might be able to make sense of why these patterns emerge.

No. of letters | No. of letters the same | No. of combinations | No. of combinations with no letters the same |

3 | 2 | 3 | 6 |

4 | 2 | 12 | 24 |

5 | 2 | 60 | 120 |

5 | 3 | 20 | 120 |

Looking at these results we can see that the number of combinations for each name are half that of the number of combinations when there are no letters the same. This works fine up until when, on closer inspection, this only works for the names that have two letters the same. So when that original theory is used on a name that has more than two letters the same it doesn't work. The obvious and simple answer that comes to mind is to divide the number of combinations when no letters are the same by the number of letters the same. This would work up until we come to the word with three letters the same and it doesn't work. Working on the basis that the number of combinations when no letters are the same will have to be divided to come up with

Conclusion

N! / (A!)(B!)

This follows the same method as the previous equation. In the previous there was only one letter that repeated, but in the case of a word that has more than one letter that repeats, you still divide by the number of letters repeated, but you also divide by the other letter that repeats as well. So A is the first letter that repeats more than once and B is the second letter that repeats more than once. So effectively the longer the word goes on and has letters that repeat, the longer the equation will be. However many letters there are that repeat, there will be that amount of brackets.

This can be tested and verified using the combinations that I worked out previously for the word XXXYY:

XXXYY

XXYXY

XXYXX

XYXXY

XYXYX

XYYXX

YXXXY

YXXYX

YXYXX

YYXXX

There would usually be 120 combinations. But, as there are letters the same it will be reduced. Using the formula N! / (A!)(B!) I can work out that there will only be 10 different combinations. The original number, 120, has been divide by 12 to get the number of combinations for this word and 12 = (3*2)(2*1), this proves that the formula works. The reason that there are this number of combinations is that, as explained earlier in my investigation, there will be lots of combinations that will be exactly the same as each other.

Now that I have worked out these equations I can work out the number of combinations for absolutely any word:

MATHEMATICS = (11*10*9*8*7*6*5*4*3*2*1)

(2*1)(2*1)

= 9979200

HYPOTENUSE = (10*9*8*7*6*5*4*3*2*1)

(2*1)

= 1814400

FACTORIAL = (9*8*7*6*5*4*3*2*1)

(2*1)

= 181440

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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