EMMA'S DILEMMA
EMMA'S DILEMMA INTRODUCTION: I will begin my Investigation by finding the number of arrangements does the word LUCY, EMMA and other words have. By doing this I will come up with a formula, which I could find, any number of arrangements, repeated words or no repeated words. I will begin by finding the total arrangement does the word LUCY have. Then I will find the arrangement for a 1-lettered word, 2-letered word, 3-lettered word and then 5-lettered word, with no repeated letters. After that I will look at the arrangements and see if there is a pattern to the arrangements and find a formula, which can find any arrangements with no letter repeated. Part 1 I will start my investigation by finding the number of arrangements does the word LUCY have. . LUCY 7. ULCY 13. CULY 19. YLUC 2. LUYC 8. ULYC 14. CUYL 20. YLCU 3. LCUY 9. UYLC 15. CLUY 21. YULC 4. LCYU 10. UYCL 16. CLYU 22. YUCL 5. LYCU 11. UCLY 17. CYLU 23. YCLU 6. LYUC 12. UCYL 18. CYUL 24. YCUL I have found 24 arrangements in the word LUCY. Now I am going to find out how many number of arrangements does a three-letter word, two letter word and one letter word have. -lettered word: .A I have found only one arrangement in a one-letter word. 2-lettered word: .CA 2.AC I have found only two arrangements in a two-lettered word. 3-lettered word: . CAN 3. ACN 5.
Emma's Dilemma
Emma's Dilemma Arrangements for Emma: emma emam eamm mmae mmea meam mema mame maem amme amem aemm There are 12 possibilities; note that there are 4 total letters and 3 different. What if they were all different like Lucy? Arrangements for Lucy: lucy ucyl cylu ycul luyc ucly cyul yclu lcuy ulcy culy yulc lcyu ulyc cuyl yucl lyuc uycl clyu ylcu lycu uylc cluy yluc There are 24 different possibilities in this arrangement of 4 letters all different. Double the amount as before with Emma's name, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For example there are 6 arrangements with Lucy beginning with L, and 6 beginning with u and so on. 6 X 4 (the amount of letters) gives 24. What if there were 4 letters with 2 different? Arrangements for aabb: aabb abab baab abba baba bbaa There are 6 arrangements for aabb. From 2 different letters to all different letters in a 4-letter word I have found a pattern of 6, 12 and 24. As it is easier to see what is happening with more difficult arrangements I will do a table for more letters and try and look for a more meaningful explanation. What if there was a five-letter word? How many different arrangements would there be for that? As I have found that there were 24 arrangements for a 4 letter word with all different
Emma's dilemma.
We were given the name 'Emma' and asked to look at it and see how many different ways the letters could be arranged in. But 'Emma' has a repeat in it; it has two m's. I decided to start off with the simpler option by not having any repeats in the words, also I thought I would go through first doing a 2 letter word then 3 and then 4... and then I would put all the information into a table then try and work out repeats. JO - jo - oj SAM - sam sam - asm I put them into a bit more sma - mas order so it was easier to understand. ams - msa I did this by doing the 's' first asm - ams then the 'a' and then the 'm'. msa - sma mas TONI - toni - nito - toin - niot - tion - ntio <-- There are 6 different arrangements for each letter. - tino - ntoi So altogether for a 4 letter word there are 24 - tnoi - noti different ways. - tnio - noit ------- ------- - onti - itno - onit - iton - otni - iotn - otin - iont - oitn - into - oint - inot KIREN - kiren - keinr - irenk - ierkn - ikern - kirne - keirn - irekn --------- - iknre - kienr - kerin - irnek - inker - ikner - kiern - kerni - irnke - inkre There are 24 different - kinre --------- - irken - inrkr ways for 'i' too. So - kiner - knire - irkne - inrek altogether there are --------- - knier --------- - inekr 120 different ways
Emma's Dilemma
Emma's Dilemma Emma and Lucy are playing with arrangements of the letters of their names. Part 1 Investigate the number of different arrangements of the letters of Lucy's name. Part 2 Investigate the number of different arrangements of the letters of Emma's name. Part 3 Investigate the number of different arrangements of various groups of letters. Emma's Dilemma Emma and Lucy were playing with arrangements of the letters of their names. Part 1 My first task was to investigate the number of different arrangements for the letters in Lucy's name. I began by listing the arrangements, one by one, following a certain structure (a structure which will hereafter be used in similar situations throughout the rest of this investigation). This started with me dividing the name "Lucy" up into its individual parts i.e. the letters L, U, C and Y. By re-arranging these, I was then able to find new arrangements. Starting with LUCY (obviously), I moved on to find LUYC (by exchanging the two letters at the end). When I found that there were no longer any arrangements starting with LU, I simply replaced the second letter with the subsequent letter and started to look for arrangements starting with LC. When the arrangements starting with L had run out altogether, I took the next letter i.e. U and started to look for arrangements starting with that. In the end, I found that there were 24
Emma's Dilemma.
Emma's Dilemma Introduction This coursework is about rearrangements of the letters in peoples names. We will be looking at how many combinations you can make out of the letters in different size names. I will then work out a formula that will let me work out the number of combinations in a name given the number of letters in that name. I will work this out in a systematic way, by changing one letter at a time and moving down the word. The names I will be using are: 2 letter name BO 3 letter name SAM 4 letter name MIKE 5 letter name KARIN Results 2 letter name BO OB For this name, there are only 2 possible combinations 3 letter name SAM ASM MSA SMA AMS MAS For the 3 letter name, there are 6 possible ways. 4 letter name MIKE IMKE KMIE EMIK MIEK IMEK KMEI EMKI MKIE IEMK KIEM EIKM MKEI IEKM KIME EIMK MEIK IKEM KEIM EKMI MEKI IKME KEMI EKIM 24 different combinations are possible with a 4 letter name 5 letter name KARIN KRAIN KINAR KNARI KARNI KRANI KINRA KNAIR KAINR KRNIA KIARN KNRAI KAIRN KRNAI KIANR KNRIA KANRI KRINA KIRAN KNIRA KANIR KRIAN KIRNA KNIAR ARINK AINKR ANKRI AKRIN ARIKN AINRK ANKIR AKRNI ARKIN AIKRN ANIKR AKRNI ARKNI AIKNR ANIRK AKINR ARNKI AIRKN ANRKI AKNIR ARNIK AIRNK ANRIK AKNRI RINKA RNKAI RKAIN RAKIN RINAK RNKIA RKANI RAKNI RIKAN RNAKI RKINA RANIK
Emma's Dilemma
Emma's Dilemma - Mathematics Coursework Scott McMenemy * Emma and Lucy are playing with the different arrangements of the letters of their names. There are a number of different arrangements in the letters of Lucy's name: LUCY ULCY CYUL YLUC LUYC ULYC CYLU YLCU LYUC UYLC CLUY YCLU LYCU UYCL CLYU YCUL LCYU UCLY CULY YUCL LCUY UCYL CUYL YULC There are an overall 24 different arrangements for the letter in Lucy's name. * There are a number of different arrangements for the letters od Emma's name also: EMMA MEMA AMME EMAM MEAM AMEM EAMM MMEA AEMM MAME MAEM MMAE There are an overall 12 different arrangements for the letters in Emma's name. * I have discovered that there are 24 arrangements for any 4 letter word as long as all letters are different - e.g. Lucy, Fred, John. Taking n as the number of letters in the word, we can say that the number of arrangements for a word which has all different letters in n factorial, or n! We can use the name FRED as an example. The numbers of arrangements are shown below: FRED RFED EFRD DFRE FRDE RFDE EFDR DFER FDRE RDFE EDFR DRFE FDER RDEF EDRF DREF FEDR REDF ERDF DEFR FERD REFD ERFD DERF Factorial is used because for the first choice letter of the word, there are 'n' choices, and for the second letter there are 'n-1' choice, and for the
Emma's Dilemma.
Emma's Dilemma We are investigating the number of different arrangements of letters. Firstly we arrange Emma's name... EAMM EMAM MEMA MEAM MMEA MMAE MAEM MAME AMME AEMM AMEM EMMA There are 12 different arrangements for the name Emma. Note that there are four letters in total and three different letters. Next I arranged Lucy... Lucy Ulcy Ucyl Cuyl Ycul Luyc Ucly Cyul Yluc Yluc Lycu Uycl Culy Yucl Lcuy Ulyc Cylu Yclu Lcyu Uylc Clyu Ylcu There are 24 different possibilities in this arrangement of the name Lucy with four letters that are all different. Twice as many as before in Emma's name which had four letters but three different letters. In the name Lucy I noticed that there were six possibilities beginning with a different letter each time. For example in the name Lucy there were six arrangements starting with L six starting with U six starting with C and six starting with Y. Therefore I found that 6 times the amount of letters (in this case 4) equaled the amount of different arrangements for the name. If we take aabb as a further example for the amount of arrangements, what would happen? Arrangements for aabb... aabb baab baba abab abba bbaa There are six different arrangements for the letters aabb as the table shows. From Emma with two different letters to Lucy with all different letters I have found a pattern for
Emma's Dilemma
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
Emma's Dilemma
Maths Investigation: Emma's Dilemma In my investigation I will investigate the number of different combinations a word can be put in. For example the word... Tim. The letters in this word can be mixed up to show all the possible variations of combinations the letters can be put in. So a variation of the name Tim would be... Mit. E.g. TIM, ITM, MIT, TMI, ITM, MTI. ...this shows all the possible combinations the letters can be put into. A total of 6 different combinations can be achieved. I will begin by investigating the name LUCY. I will work out all the possible letter combinations that can be produced from this name. I have chosen this name because it has no letters the same and I first intend to investigate words with no letters repeated before perhaps moving on to that situation. Total = 24 variations. I will try the same with a 3-letter name to see if there is some sort of pattern. Total = 6 variations I can not yet see any sort of connection yet other than they are both even numbers, I will do the same thing with a 2-letter name. Total = 2 different arrangements. I will now draw a table to show my results this may help me find connection more easily because the links will be more visible. Table of Results Number of Letters Number of Different Arrangements 2 2 3 6 4 24 From the first two results the amount of
Emma's Dilemma.
Emma's Dilemma Firstly we arrange EMMA's Name. )EAMM 7)MAEM 2)EMAM 8)MAME 3)MEMA 9)AMME 4)MEAM 10)AEMM 5)MMEA 11)AMEM 6)MMAE 12)EMMA Secondlywe arrange lucy's name. )Lucy 12)Cyul 22)Yulc 2)Luyc 13)Culy 23)Ycul 4)Lycu 14)Culy 24)Yluc 5)Lcuy 15)Cylu 25)Ucyl 6)Lcyu 16)Clyu 7)Ulcy 17)Cuyl 8)Ucly 18)Yluc 9)Uycl 19)Yucl 0)Ulyc 20)Yclu 1)Uylc 21)Ylcu From these 2 investigation I worked out a method: Step1: 1234---Do the last two number first then you get 1243. 243---Do the last three numbers and try the possibility. 1423. 1432. 1342. 1324, because the number 2 has been the first number of last three numbers, so we don't do it again. Step2: we have list all arrangements of 1 go front, so we do 2 go front. 2134 and we do same thing to it, it will like this: 2134---2143, 2143---2431,2413,2314,2341 Step3: We have finished 2 go first, then let's do 3 go ahead. 3124---3142, 3142---3241,3214,3412,3421 Step4: We have finished 3 go ahead, then try 4 4123---4132, 4132---4231,4213,4312,4321 We have list all arrangement of 1234, use this method we can arrange the number which has 5 figures or more. We are trying to work out a formula which can calculate the number of arrangement when we look at a number. Let's list all the arrangment for 1234: 234 4123 243 4132 324 --- 6 arrangment 4231 ---- 6 arrangement 342 4213 432 4321 423 4312 2134 3124 2143 3143 2341