Emma's Dilemma
MOHAMED KHALIFA MATHEMATICS INTRODUCTION: For mathematics, we were given the Coursework of investigating "Emma's Dilemma." This Coursework is set up in a number of steps which builds up to a formula in which one could enter any number of letters, with or without repeated letters, and come out with the number of possible permutations in which these words could be expressed. For this reason, I decided that I will not follow the steps and begin by investigating what is required, which is come up with the formula. I shall begin my task by looking at the possible number of ways in which I could write a word which has no letters repeated in it. I shall express the phrase "possible number of words" with the symbol ?. WORDS WITH NO REPEATED LETTERS: I shall start by looking at such words starting from single lettered words (Most of my words will not make sense): - ? (a) = 1 2- ? (ed) = ed, de = 2 3- ? (cat) = cat, cta, tac, tca, act, atc = 6 4- ? (lucy) = (lucy, luyc, lcuy, lcyu, lycu, lyuc) * 4 = 24. Notice how I multiplied the last example by 4 to get the ?. This is because I have noticed something as I was working out the ?. I have found that I was going to repeat everything I was doing 4 times. Similarly, I have realised that I was doing every thing 3 times for the second letter, twice for my third letter and once for my last. Hence, I could say that to work out the
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 ) I will methodically write down all the combinations of the letters of Emma's name:- MEMA EMMA AMME MEAM EMAM AEMM MAEM EAMM AEMM MAME MMAE MMEA As you can see from the above the results are: 4 letters, 2 the same, 12 combinations. (Remember there are two letters that are the same and therefore this affects the amount of combinations, as we will see later). 2) With long names it is hard to work out the number of different combinations and ideally we need to find a formula instead of having to right them all out. Firstly, it would probably be easier if we started with a simple name with no repeating letters, not to complicate things, and then I will simply work up: Name: AD Two letters AD DA Combinations: 2 Name: JON Three letters JON OJN NJO JNO ONJ NOJ Combinations: 6 Name: LUCY Four letters LUCY UCLY YLCU CULY LUYC UCYL YLCU CUYL LYUC ULCY YUCL CLYU LYCU ULYC YULC CLUY LCUY UYLC YCLU CYUL LCYU UYCL YCUL CYLU Combinations: 24 Name: SIMON Five letters SIMON SOIMN SMION SNIOM SIMNO SOINM SMINO SNIMO SINMO SONMI SMNOI SNMIO SIOMN SONIM SMNIO SNMOI SINOM SOMIN SMOIN SNOMI SIONM SOMNI SMONI SNOIM In this case, the name I chose was too long but with 24 combinations with 'S' first, there are 120 combinations overall (24x5). Combinations: 120 From the above I
Emma's Dilemma.
Emma's Dilemma I am trying to find out the different arrangements of letters in a word with so many letters. Also I am going to try and find out a way of finding these mathematically and also to try and find a formula which will work with my theory. My way of working these out is by a simple method. First I used the name Lucy, this is because all the letters in this name are different. So it will be easier. Then we used this method of: ) I would keep the first letter the same all the time. 2) Then I would keep the second letter the same. 3) Then I would rotate the other letters around to make different arrangements. 4) After I have done this I would change the second letter to a different in the word but not the first letter. 5) Then I would do the same as in step 3 where I would rotate the last 2 letters, such as below. 6) Do the same but with the same first letter but a different second letter that you haven't already used. 7) Once I have done this I have found all the different arrangements with the first letter. Then I have to change the first letter, such as below, and then do the same method as before with all the different first letters. This is all the different arrangements for a 4 lettered word with all the letters different. LUCY LUYC LYCU LYUC LCYU LCUY UCYL UCLY ULCY ULYC UYLC UYCL CYLU CYUL CUYL CULY CLUY CLYU YCLU YCUL YULC
Emma's Dillemma
. Emma 2 Emam 3 Eamm 4 Aemm 5 Amem 6 Amme 7 Meam 8 Mema 9 Mame 0 Maem 1 Mmae 2 Mmea There are 12 different arrangements of Emma 2. Lucy 2 Luyc 3 Lyuc 4 Lycu 5 Lcuy 6 Lcyu 7 Culy 8 Cuyl 9 Cylu 0 Cyul 1 Cluy 2 Clyu 3 Yclu 4 Ycul 5 Ylcu 6 Yluc 7 Yucl 8 Yulc 9 Ulcy 20 Ulyc 21 Uylc 22 Uycl 23 Uclu 24 Ucul There are 24 different arrangements of Lucy 3. Jo 2 Oj Joe 2 Jeo 3 Eoj 4 Ejo 5 Oje 6 Oej Ann 2 Nan 3 Nna JJ Number of Letters With Repeated Letters No Repeated Letters JJ Jo 2 2 Ann Joe 3 3 6 Emma Lucy 4 2 24 The "no repeated letters" field is equal to the "With repeated letters" field but doubled i.e Lucy = 4 letters = 24 arrangements all you need to do then is to divide it by 2 and you will have the value of "With repeated letters" field and visa versa. There is a quicker solution to this problem. All that you need to do is (If you need a 4 letter word) you would do 1x2x3x4. If you needed 5 letters you would do 1x2x3x4x5 and so on. If you want to find out the value of the repeated letters field you would then just divide it by 2. I predict that a 5 letter word will have 120 combinations because 1x2x3x4x5 = 120. 20 combinations will be too much so I will do a 5-letter word that has a repeated letter such as Barry Larry or Annie. This will, according to my theory
Emma’s Dilemma
Mathan Navaratnam 11F 24.10.2001 Mathematics Coursework Emma's Dilemma The different arrangements for the name Emma are as follows: Emma emam eamm mmae Mmea meam mema mame Maem amme amem aemm There were a total of twelve combinations of the rearrangement of the name Emma. The name Emma consists of four letters where three of them being different, the two like letters being m. Emma's friend however, Lucy, name consists again of four letters but this time all the letters are different. The number of arrangements for the name Lucy is as follows: Lucy ucyl cylu ycul Luyc ucly cyul yclu Lcuy ulcy culy yulc Lcyu ulyc cuyl yucl Lyuc uycl clyu ylcu Lycu uylc cluy yluc This time there were twenty-four different possibilities in rearranging the name Lucy. This is significant, as it is exactly double the number of arrangements of the name Emma where there were 4 letters but where only 3 were different. Looking back at the table above I think it would be useful to note how to predict the number of arrangements. So far I have worked out for names with four letters. If I am to use the name Lucy then I notice in my above table that there are six combinations for the rearrangements if the rearrangement was to say the letter l, six combinations with rearrangements starting with the letter u and so forth. Thus 4 (number of different letters) multiplied by 6 (number of
Emma's dilemma
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. ------------------------------------------------------------------------------------------------------- --The different arrangements of EMMA's name: EMMA MMAE MAEM MEAM AEMM EAMM EMAM MMEA MAME MEMA AMME AMEM There are 12 different combinations of rearranging the letters of EMMA. -------------------------------------------------------------------------------------------------------The different arrangements of LUCY's Name: LUCY ULCY CLUY LCUY LYUC YLUC LUYC ULYC CLYU LCYU LYCU YLCU CYLU YCLU UCYL UYCL CUYL YULC CYUL YCUL UCLY UYLC CULY YUCL There are 24 different combinations of rearranging the letters of LUCY. ------------------------------------------------------------------------------------------------------- The method, which I used in order to find the possible solutions for both 'LUCY' and 'EMMA', was the same. I first took all the possible first 2 letters of the name: e.g for EMMA it was- EM, EA, ME, MM, MA, AE, AM. The way in which I was able to get all the possible first 2
Emma's Dilemma.
Maths Coursework Emma's Dilemma Investigate the number of different arrangements of the letters of Emma's name Introduction: For this piece of coursework I am going to investigate the number of arrangements in a word such as: BIT. For the word BIT BIT is an arrangement and TIB is another. After that I will look at words with different numbers of letters to see if there's a pattern emerging. I will use this pattern to work out some algebraic equations for the number of arrangements in a word such as a 20 letter word. After that I will investigate words with a double letter like : EMMA, to see if there's pattern for that. I will work out the equation for that and then I will extend the investigation even further. Part 1: Words with no doubles 2 letter word I have found there are two combinations for a 2 letter word. The equation is 1x2 because there are 2 combinations of letters that can be assigned to the 1st letter and then once you've assigned a letter to the 1st letter there's only 1 letter that can be assigned as the 2nd letter. x2=2 combinations 3 Letter word I have found there are 6 combinations for a 3 letter word. The equation is 1x2x3 because there are 3 combinations of letters that can be assigned as the 1st letter, then there are 2 combinations of letters that can be assigned as the 2nd letter and finally there's
Emma's Dilemma.
Emma's Dilemma By Lucy Ellery 11RK Teacher: Mr. Crowther Aim: The aim of this investigation is to find out a general formula for the arrangements of different names. I want to find the number of different arrangements for the name 'EMMA'. EMMA EMAM EAMM, AEMM AMEM AMME, MEAM MAEM MEMA MAME MMEA MMAE There are 12 arrangements for the name 'EMMA'. Now I will find the number of arrangements for the name 'LUCY', which has the same number of letters. I predict that 'LUCY' will also have 12 variations, because it has the same amount of letters. LUCY LCYU LCUY LUYC LYCU LYUC, ULCY ULYC UCLY UCYL UYCL UYLC, CLUY CLYU CYLU CYUL CULY CUYL, YCLU YCUL YLCU YLUC YUCL YULC There are 24 variations for the name 'LUCY'. I have seen that although 'LUCY' has the same amount of letters as 'EMMA', it has double the variation of arrangements. I think that this may be because 'EMMA' has a double letter in it. I will now take some new names and see what results they give me. 1 lettered name- J -one variation 2 lettered name- JO, OJ-two variations 3 lettered name- JOE, JEO, OJE, OEJ, EJO, EOJ-six variations. Now I will draw a table with my results and see if I can see any patterns. Number of Letters Number of Permutations 2 2=(1x2) 3 6=(1x2x3) 4 24=(1x2x3x4) 5 20=(1x2x3x4x5) My prediction for a name with five letters was made because looking at the table I saw that as you
Compare and contrast the fictional letters in 'Birdsong' with the real letters written to Vera Brittain by Edward Brittain and Roland Leighton. Which do you find more moving and powerful and why?
Compare and contrast the fictional letters in 'Birdsong' with the real letters written to Vera Brittain by Edward Brittain and Roland Leighton. Which do you find more moving and powerful and why? The letters sent in the war were one of the most important ways of communicating with loved ones. A stereotypical view would be that letters would seem quite heart-felt and filled with detailed descriptions of the reality thrown at the soldiers. However in reality the emotion doesn't come across frequently. In 'Birdsong,' Faulks picks out key elements to format emotive letters. The tone of each letter varies, but a general positive attitude comes across through the letters, with the exception of 'Stephen's' letter. The tone in 'E.Brittain's' and 'Leighton's' letters did not make much of an impression; it was a detailed account of their normality in the war. Their viewpoint of the war did not seem to emerge. Both letters consists of war jargon and formal language which prevents their true feelings to be empathised with, making their letters less moving than 'Faulk's' fictional letters. In contrast with the reality, the readers can vaguely pick up on the characters' viewpoint in 'Birdsong' through the language and tone making the letters to appear more powerful. 'Faulks' has intentionally written the character's letters before the Battle of the Somme. In 'Weir's' letter the first