Emma's Dilemma
) This is all the different arrangements of the name EMMA. EMMA EAMM EMAM MEMA MMEA MAME MAEM AEMM AMME AMEM MMAE MEAM 2) This is all the different arrangements of the name LUCY LUCY LCUY LYUC LUYC LCYU LYCU ULCY ULYC UCLY UCYL UYCL UYLC CLUY CLYU CYLU CYUL YLCU YLUC YULC YUCL YCLU YCUL CULY CUYL The reason EMMA (a 4 letter name) has only 12 different arrangements LUCY has 24 is because in EMMA there are two of the same letters in the name and so it restricts the possible amount of combinations, however in the name LUCY all the letters in the names are different so you get the maximum amount of combinations. 3) These are the names that I chose to investigate for different arrangements. I chose 3 names under each amount of letters. 2 Letter Names This is all the different arrangements of the name JO JO OJ This is all the different arrangements of the name MO MO OM This is all the different arrangements of the name TY TY YT 3 Letter names This is all the different arrangements of the name JIM JIM JMI MIJ IMJ IJM MJI This is all the different arrangements of the name BEN BEN NEB BNE ENB EBN NBE This is all the different arrangements of the name ANN. (I chose Ann because it had 2 of the same letters in its name, and to see what effect it had on the amount of different
Emma's Dilemma.
Maths Coursework 2002 Emma's Dilemma Rory O'Connell The Aim of this investigation is to see how many combinations of letters there are in names and other letter combinations. For example the name Emma has the following combinations: EMMA EAMM EMAM AMEM AMME AEMM MEAM MAEM MEMA MAME MMEA MMAE In the word EMMA there are 12 possible combinations These are the possible combinations for the word LUCY: LUCY UYLC LUYC UYCL LCUY UCYL LCYU UCLY LYUC LYCU CLYU CLYU CULY CUYL CYUL CYLU YLUC YLCU YCUL YCLU YULC YUCL ULCY ULYC In the word LUCY there are 24 possible combinations. Twice the amount of arrangements in the word EMMA, despite having the same amount of letters. I then looked at the number of combinations of letters there were in names of varying length: JO SAM FRED DARYL GERALD I did this by writing out all the possible combinations for each name, For example: SAM SMA AMS ASM MAS MSA And JO OJ Results Table Name Number Of Combinations JO 2 SAM 6 FRED 24 LUCY 24 EMMA 2 DARYL 20 GERALD 720 I have found an equation, which will tell you the number of letter combinations in each word (except EMMA), it is based on the following idea: In the word FRED for example, there are four letters. When rearranging the letters, there are four possibilities for where the first letter could be
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
Emma's Dilemma.
G.C.S.E Maths Coursework Emma's Dilemma For my G.C.S.E maths coursework I have been given the task to investigate how many times the letters of any name can be arranged. The method I am going to be using is to explore how many times the letters of short and long names can be arranged. Once I have got the hang of this I will investigate how many times short and long letter names with repeated letters can be arranged. Once I have completed this, I will create a general formula which could be used with any name with or without repeated letters. AJ JA How many arrangements can be made from a name with two different letters like AJ? AJ has a total of two arrangements. How many times can a three letter name like JON be arranged? Here are the arrangements:- JON NJO JNO NOJ OJN ONJ From this table, we can see that there are a total of 6 arrangements. EMMA EAMM EMAM MEAM MEMA MAEM MMEA MAME MMAE AEMM AMEM AMME Below is a table showing the arrangements for EMMA. This time there are 2 repeated letters. Let's see how many times this name can be arranged:- For EMMA, there are 12 possible arrangements. There are 2 different letters and two letters the same, and there is a total of 4 letters. Every time I move the letter E along there are 3 possibilities. Also, if I multiply the four vertical rows by the three horizontal rows of the table, (4 x 3) I will get an
Emma's Dilemma
Emma is playing with the different arrangements of the letters in her name, here is a list of all the different arrangements: - emma 2- emam 3- eamm 4- amme 5- amem 6- aemm 7- mmae 8- mmea 9- mema 0- mame 1- maem 2- meam To make sure that she listed all of the possible arrangements, Emma used a systematic formula. Lets try it with the letters abcd: -abcd first start off by keeping the first two letters the same and then swapping the last two letters. 2-abdc Then keep the first letter where it is and swap all the second number with the third. 3-acbd Now you can swap the last two numbers and keep the first two where they are. 4-acdb Next swap the second letter with the only letter that has not been second yet which should be D, then you can swap the last two letters again. 5-adbc 6-abdc Now that you have listed all of the arrangements for the letter A you can repeat this process but each time swap the first letter for one of the others. 7-bacd 8-badc 9-bcad 0-bcda 1-bdac 2-bdca 3-cabd 4-cadb 5-cbad 6-cbda 7-cdab 8-cdba 9-dabc 20-dacb 21-dbac 22-dbca 23-dcab 24-dcba Emma then tried to find the different arrangements of her friend Lucy's name, she used the same systamatic formula. - lucy 2- luyc 3- lcuy 4- lcyu 5- lycu 6- lyuc 7- ulcy 8- ulyc 9- ucyl 0- ucly 1- uycl 2- uylc 3- culy 4- cuyl 5- clyu 6- cluy 7- cyul 8-
Emma's Dilemma
Danny Soopen GCSE Maths Project - "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. Part 1: Ways of arranging EMMA's name: EMMA MEMA MMAE AMME EMAM MEAM MAEM AMEM EAMM MMEA MAME AEMM For each new beginning letter, there are 3 different possible combinations. There are a total of 12 possible different combinations. Part 2: Ways of arranging LUCY's name: LUCY ULCY CULY YUCL LUYC ULYC CUYL YULC LCUY UCLY CLUY YCUL LCYU UCYL CLYU YCLU LYUC UYLC CYLU YLUC LYCU UYCL CYUL YLCU For each new beginning letter, there are 6 different possible combinations. There are a total of 24 possible different combinations. This combination consists of all the letters being different. I will also try and find a formula for this arrangement, and number of letters. Once I have discovered these formulae I am going to investigate, other combinations of letters and different amounts of letters. I will then try to discover a link between the formulae to enable me to find a formula for the general case. There are 24 different possibilities in
Emma's Dilemma
Maths Coursework Emma's Dilemma Emma's Dilemma The name Lucy has got four letters with none of the letters repeated. This means that when I come to choosing a letter I have a choice of four then 3 then 2 then 1 if I write those out and multiply them together I will get the number of arrangements. Which is 4 x 3 x 2 x 1, which also can also be written as 4! In my coursework I am going to show the different arrangements of Lucy's name and I am going to find out how many how many arrangements there can be for any name even if the name has more than one letter that is repeated. I will do this by using a formula that I am going to find. Here is a list of all of the arrangements of Lucy's name: . LUCY 2. LUYC 3. LCUY 4. LCYU 5. LYUC 6. LYCU 7. ULYC 8. UCLY 9. ULCY 0. UCYL 1. UYCL 2. UYLC 3. CYLU 4. CYUL 5. CULY 6. CUYL 7. CLUY 8. CLYU 9. YLUC 20. YLCU 21. YCUL 22. YCLU 23. YULC 24. YUCL There are 24 different arrangements of Lucy because first I had 4 to choose form then I had 3 to choose from then I had 2 to choose from then I had 1 to choose from. So if I times 4 x 3 x 2 x 1 I will get 24. This is the same as 4! This sign is called factorial. And this is another way to find out how many arrangements there are in Lucy's name, this method is called factorial. With this method I do not have to list all of the different ways of Lucy's
Emma's Dilemma
EMMA'S DILEMMA 4 LETTER NAME WITH 2 LETTER THE SAME EMMA ) EMMA 2) EMAM 3) EAMM 4) MEMA 5) MEAM 6) MMEA 7) MMAE 8) MAME 9) MAEM 0) AEMM 1) AMEM 2) AMME 4 LETTER NAME LUCY ) LUCY 2) LCUY 3) LCYU 4) LUYC 5) LYUC 6) LYCU 7) ULCY 8) ULYC 9) UCLY 0) UCYL 1) UYCL 2) UYLC 3) CYLU 4) CYUL 5) CUYL 6) CULY 7) CLUY 8) CLYU 9) YLCU 20) YLUC 21) YULC 22) YUCL 23) YCLU 24) YCUL I am now going to test this out on different names and see if I can find a pattern. 2 LETTER NAME MO ) MO 2) OM 3 LETTER NAME JOE ) JOE 2) JEO 3) EJO 4) EOJ 5) OEJ 6) OJE 5 LETTER NAME RICKY ) RICKY 2) RICYK 3) RIKCY 4) RIKYC 5) RIYCK 6) RIYKC 7) RCIKY 8) RCIYK 9) RCYIK 0) RCYKI 1) RCKIY 2) RCKYI 3) RKYCI 4) RKYIC 5) RKICY 6) RKIYC 7) RKCYI 8) RKCIY 9) RYICK 20) RYIKC 21) RYKCI 22) RYKIC 23) RYCKI 24) RYCIK 25) ICKYR 26) ICKRY 27) ICRYK 28) ICRKY 29) ICYRK 30) ICYKR 31) IKYCR 32) IKYRC 33) IKCYR 34) IKCRY 35) IKRCY 36) IKRYC 37) IYRKC 38) IYRCK 39) IYCRK 40) IYCKR 41) IYKRC 42) IYKCR 43) IRCKY 44) IRCYK 45) IRKCY 46) IRKYC 47) IRYCK 48) IRYKC 49) CKYRI 50) CKYIR 51) CKRIY 52) CKRYI 53) CKIRY 54)
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 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