Emma's dillemma

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 =

ABC, ACB, BAC, BCA, CAB and CBA.

“Lucy” is a four letter word that has all the letters different in it. By working out how many permutations of the word “Lucy” I will be able to notice a pattern between words that have all the letters different in them. This will help me with my investigation as it will help me work out the pattern between words with all the letters different, and so I will be able to find out a word that has many letters in it without a problem as long as they are all different.

The word “Lucy” is a four letter word and so the number of permutations of the word is 24 =

lucy, luyc, lcuy, lcyu, lycu, lyuc, ulcy, ulyc, ucly, ucyl, uycl, uylc, culy, cuyl, cluy,

clyu, cylu, cyul, yucl, yulc, ycul, yclu, ylcu, yluc.

I have noticed a pattern for words that are different, by finding out the permutations of words that are different such as A, AB, ABC and LUCY. I will explain what I have found out below in a formula.

For a one letter word there is only 1 permutation.

For a two letter word with all the letters different there are 2 permutations.

For a three letter word with all the letters different there are 6 permutations.

And for a four letter word with all the letters different there are 24 permutations.

I will now put this into a formula which will be easy to work out the pattern with. I will use (n) to indicate the number of letters in the word, (p) to indicate the permutations.

Formula showing how to work out the permutations of words that have all the letters different:

(n) Number of letters in word (p) Permutations of word

1 1 = 1

2 2 = 1×2

3 ...

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(n) Number of letters in word (p) Permutations of word

1 1 = 1

2 2 = 1×2

3 6 = 1×2×3

4 24 = 1×2×3×4

By looking at this formula I have seen a pattern emerging which shows that the amount of permutations in a word that has all the letters different, depends on how many letters it has. I have noticed that the amount of letters represent the factorial and thus makes it easy to work out the amount of permutations of the word. Factorial of any number is that number multiplied by its preceding numbers from the number 1 such as a 4 letter word with all letters different will be 1×2×3×4. For example, “AB” is a 2 letter word which means that with my prediction its factorial is 2, “AB” is 1×2 = 2 permutations of the word. This appears to be right as AB does have 2 permutations.

Example:

In order to test the pattern I will now use a 5 letter word with all the letters different. The word I am going to use will be “Wayne” it doesn’t matter what word I am using as long as all the letters are different.

In the word “Wayne” I expect there to be 120 permutations because there are 5 letters in the word and they are all different so 5 Factorial =1×2×3×4×5 = 120

wayne wayen wanye waney waeny waeyn wyane wyaen wynae wynea wyena wyean

wnyae wnyea wnaye wnaey wneay wneya weyna weyan wenya wenay weany weayn

awyne awyen awnye awney aweny aweyn aywne aywen aynwe aynew ayenw ayewn

anywe anyew anwye anwey anewy aneyw aeynw aeywn aenyw aenwy aewny aewyn

yawne yawen yanwe yanew yaenw yaewn ywane ywaen ywnae ywnea ywena ywean

ynwae ynwea ynawe ynaew yneaw ynewa yewna yewan yenwa yenaw yeanw yeawn

naywe nayew nawye nawey naewy naeyw nyawe nyaew nywae nywea nyewa nyeaw

nwyae nwyea nwaye nwaey nweay nweya neywa neyaw newya neway neawy neayw

eaynw eaywn eanyw eanwy eawny eawyn eyanw eyawn eynaw eynwa eywna eywan

enyaw enywa enayw enawy enway enwya ewyna ewyan ewnya ewnay ewany ewayn

Just as I expected there are 120 permutations for a 5 letter word that has all the letters different. This shows that my formula seems to have been correct, I could however test it even further by doing the 6th factorial, but I believe that I have already proved my formula or pattern to be correct, and testing it even further would take up far too much time and effort as I have already spent a lot of time proving the 5th factorial.

Conclusion:

I have come to a conclusion for letters in a word that are different and it is that if n is equal to the number of letters in a word, then n factorial will be the different number of permutations of that word.

Part 2: Now I will further my investigation by looking at words with 2 repeated letters. I will be comparing if there is a link between words with different letters and words with repeated letters. I will see if the same formula is to be applied for words with repeated letters as it was in the first part 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 less permutations in “Emma” due to the fact that two letters are repeated, because two letters are repeated it reduces the amount of permutations by half due to it having 4 letters in total and 2 of them repeated. This is what I think at the moment but continuing my investigation will let me solve the problem.

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

AA is a 2 letter word with two letters repeated and has only one permutation = AA

EGG is a 3 letter word with two letters repeated and has three permutations =

EGG, GEG, GGE

EMMA is a 4 letter word with two letters repeated and has twelve permutations =

emma, emam, eamm, mema, meam, mmea, mmae, mame, maem, amme, amem, aemm

Now that I have got these results from words that have two repeated letters I can compare it to the results I got from part 1, when I was investigating words with all the letters different. I will now make a table of it below showing the differences between “EMMA” and “LUCY” or you could say between words with two repeated letters and words with all the letters different. I will see if it will match my hypothesis I made in part 2.

Words with two repeated letters Words with all letters different

AA = 1 AB = 2

EGG = 3 ABC = 6

EMMA = 12 LUCY = 24

Conclusion:

As expected from my results it shows that my hypothesis is correct, it seems to show that there is a reduction of half due to there being 4 letters in total and 2 of them being repeated. So the formula for a word that has two repeated letters is the factorial of that word (number of letters in word) divided by number of letters repeated. Now I will be testing this theory because I still think it may have a weakness, because I have 4 letters in total and 2 of them were repeated so I automatically know that it would have been a reduction of half, but I need to see if it is the same when I have a 5 letter word and 2 of the letters are repeated, so that’s what I will be doing now.

Hypothesis: I expect there to be 60 permutations for a 5 letter word with two letters repeated, because in part 1 I got 120 permutations for a 5 letter word with all the letters different. I will use the word “SHELL” because it is an appropriate word that fits the criteria.

SHELL is a 5 letter word with two letters repeated and has 60 permutations =

shell shlel shlle sehll selhl sellh slehl slelh slhel slhle sllhe slleh hsell hslel hslle hesll

helsl hells hlesl hlels hlsel hlsle hllse hlles ehsll ehlsl ehlls eshll eslhl esllh elshl elslh

elhsl elhls ellhs ellsh lhesl lhels lhsel lhsle lhlse lhles lehsl lehls leshl leslh lelsh lelhs

lsehl lselh lshel lshle lslhe lsleh llesh llehs llseh llshe llhse llhes

From these results, I think I have enough information to assume that when you have a word with two repeated letters in it, you have to divide the factorial of the word (amount of letters in the word) by the number of letters repeated. For example, “EMMA” the factorial is 4 so you do the factorial divided by 2 which is 4!/2 = 12 which is the right answer.

Part 3

Hypothesis: Now that I have found out the pattern to words that have two repeated letters, I need to further my enquiry by finding the pattern for words that have 3 repeated letters. I think that it may have something to do with what I have done for words that have two repeated letters; it may involve the same rule which is, to divide the factorial of the word (amount of letters in the word) by the number of letters repeated. I will be finding out the permutations for a 5 letter word with 3 repeated letters i.e. AAADE. I expect there to be 40 permutations because there are 5 letters in the word and 3 repeated letters, so 5! ÷ 3 = 40

AAA is a 3 letter word with three letters repeated and has only one permutation =

AAA

AAAD is a 4 letter word with three letters repeated and has four permutations =

AAAD, AADA, ADAA, DAAA

AAADE is a 5 letter word with three letters repeated and has twenty permutations =

AAADE, AAAED, AADAE, AADEA, AAEDA, AAEAD, ADAAE, ADAEA, ADEAA, AEADA, AEAAD, AEDAA, DAAAE, DAAEA, DAEAA, DEAAA, EAADA, EAAAD, EADAA, EDAAA

I am puzzled at this result because I was expecting the result to be 40 permutations; instead I got half of that which leaves me confused. Now I will have to review what I got when I did all three parts, below is a review of my results, it shows what I got when I was undergoing my investigation.

5 letter word with all letters different 5 letter word with 2 letters same

5! = 120 5! ÷ 2 = 60

5 letter word with 3 letters same

5! ÷ 3 = 20

It shows that when all the 5 letters were different the amount of permutations was 120 because of the 5 factorial. Then it shows the amount of permutations halved due to the 5 factorial being divided by 2 repeated letters. It then shows it being divided 6 because 120 divided by 6 gives 20 permutations. This is all one way looking at it but if I rearrange the table you will be able to see the link.

5 letter word all diff letters 5 letter word 2 letters same 5 letter word 3 letters same

5! = 120 120 ÷ 2 = 60 120 ÷ 6 = 20

120 120 ÷ 2! 120 ÷ 3!

2! 3!

Now I can see a definite link, the repeated letters seem to indicate that as you keep on dividing by repeated letters, the amount of repeated letters you divide by is the amount of factorial you divide by. So if you have a 5 letter word with 2 repeated letters you will have to divide by 2 factorial, so it will be 5! ÷ 2! Which at the moment in time seems correct, however I will need to prove it and so now I will be testing this new idea on a 6 letter word with 3 repeated letters. The word I will be using will be ATAXIA.

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

ATAXIA is a 6 letter word with 3 repeated letters and it gives 120 permutations =

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).