# Emma's dillemma

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Introduction

Mathematics Coursework

Introduction: In this coursework I will be investigating “Emma’s dilemma” which is about permutations of letters. I will see how many permutations of the word “Lucy”. Then I will continue my investigation by finding the amount of permutations in a word that has 2 letters the same such as the word “Emma”. In order to fully understand the work and find patterns between the words and arrangements, I will need to structure it to show how I got from one point to the other. To further my investigation I will see how many permutations there are in various groups of words such as, words that have 3 letters the same and the rest different, and I will expand on this theory to find patterns relating to the work.

Part 1: For the first part of this investigation I will be finding out how many permutations there are in words that have all the letters different. I have decided to use A, B and C to make up a 1 letter word, 2 letter word and a 3 letter word such as A, AB and ABC.

A is a one letter word that has only one permutation = A

AB is a two letter word that has two permutations = AB and BA

ABC is a three letter word that has six permutations =

Middle

Part 2:Now I will further my investigation by looking at words with 2repeatedletters. I will be comparing if there is a link between words with differentletters and words with repeatedletters. I will see if the same formula is to be applied for words with repeatedletters as it was in the firstpart for letters that are different. I will use the word “Emma” as my starting block and I will be investigating on from there.

Hypothesis: I do, however feel that there will be lesspermutations in “Emma” due to the fact that twoletters are repeated, because twoletters are repeated it reduces the amount of permutations by half due to it having4letters in total and 2 of themrepeated. This is what I think at the moment but continuing my investigation will let me solve the problem.

As in the firstpart I will start off by using a 2letterword with the lettersrepeated, then a 3letterword with two letters repeated and a four letter word with two lettersrepeated i.e. Emma. To further my enquiry I will then do a 5letterword with twolettersrepeated to confirm my

Conclusion

I expect it to give me 6! ÷ 3! = 120 permutations

ATAXIA is a 6letterword with 3repeatedletters and it gives 120permutations =

ataxia ataxia ataxia ataiax ataaix ataaxi ataxia atxaai atxiaa atixaa atiaxa atiaax aatxia

aatxai aatixa aatiax aataix aataxi aaxtia aaxtai aaxita aaxiat aaxait aaxati aaixta aaixat

aaitxa aaitax aaiatx aaiaxt aaaxit aaaxti aaaixt aaaitx aaatix aaatxi axatia axatai axaita

axaiat axaait axaati axtaia axtaai axtiaa axitaa axiata axiaat aiaxta aiaxat aiatxa aiatax

aiaatx aiaaxt aixata aixaat aixtaa aitxaa aitaxa aitaax taaxia taaxai taaixa taaiax taaaix

taaaxi taxaia taxaai taxiaa taixaa taiaxa taiaax txaaia txaaai txaiaa txiaaa tiaxaa tiaaxa

tiaaax tixaaa xtaaia xtaaai xtaiaa xtiaaa xataia xataai xatiaa xaatia xaatai xaaita xaaiat

xaaait xaaati xaiata xaiaat xaitaa xiaata xiaaat xiataa xitaaa itaxaa itaaxa itaaax itxaaa

iatxaa iataxa iataax iaxtaa iaxata iaxaat iaaxta iaaxat iaatxa iaatax iaaatx iaaaxt ixataa

ixaata ixaaat ixtaaa

Conclusion: By testing the formula and finding out that it work I finally think that this is the right formula to work out any number of permutations; if you use this method you will be able to work it out. The method is: to find out the number of permutations in a word you do the number of letters (factorial) ÷ the number of repeated letters (factorial).

Key: (n) = number of letters

(r) = number of repeated letters

(!) = Factorial

Formula: n! = Number of permutations.

r!

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