GCSE Maths Coursework : Emma’s Dilemma
GCSE Maths Coursework : Emma's Dilemma I tried to find as many possible arrangements of letters from the name Emma as I could and found 12 different combinations (see separate sheet.) I did the same with the name Lucy and got 24 different combinations (see separate sheet.) I did this systematically to try to avoid missing any combinations and to make it easier to spot any patterns i.e. abcd abdc acbd acdb Instead of doing the rest of my investigation with names I decided to use letters i.e. abcd, to complete my investigation as I think it would be even easier to arrange them systematically and spot patterns. With just one letter I got one combination: So 1l = 1c With two letters I got two combinations: So 2l = 2c With three letters I got six combinations: So 3l = 6c And with four letters I got twenty four combinations: So 4l = 24c From this I found that the number of different combinations (c ) equals the number of letters (l ) factorial i.e. times every number together up to and including itself, so four letters = twenty four combinations = 1*2*3*4. I found this as I looked at the numbers of combinations and each number equalled itself times the number before it in the sequence. I then tried repeating letters as I had still not solved why Lucy's name made 24 different combinations while Emma's which contained one letter which was repeated twice only had 12,
Investigate all the possible combinations there are of a person's name, using only the letters in their name.
Emma's Dilemma Introduction This piece of course work was set to investigate all the possible combinations there are of a person's name, using only the letters in their name. For example Joe: The combinations are Joe Jeo Eoj Ejo Oej Oje Part 1 Lucy In this section, I will be exploring all the combinations of the name Lucy and if there is any correlation between the number of combinations and the number of letters in her name. Possible Combinations: . lucy 2. luyc 3. lcuy 4. lcyu 5. lyuc 6. lycu 7. ulcy 8. ulyc 9. uylc 0. uycl 1. ucly 2. ucyl 3. cluy 4. clyu 5. culy 6. cuyl 7. cylu 8. cyul 9. yluc 20. ylcu 21. yucl 22. yulc 23. yclu 24. ycul Method Firstly you choose one letters to start with, then change the second letter, for the example of Lucy we would have Lucy Lucy <-- The U stays the same but the last two letters change, this pattern is repeated. From these results I know that a 4 letter name, with no repeated letters gives 24 possible combinations, my initial thoughts where that you simply times the number of letters by 6 to get the number of possible combinations. I will be investigating this is Part 3 Part 2 Emma In this section I will be exploring the possible combinations of a name that includes a double letter. Possible Combinations: . emma 2. emam 3. eamm 4. meam 5. maem 6. mame 7. mema 8. mmea 9.
GCSE Mathematics Coursework - Emma's Dilemma
GCSE Mathematics Coursework - Emma's Dilemma Method: First, I tested different arrangements of the name 'EMMA', by systematically rearranging the letters in the name such as: -EMMA 4-AEMM 7-MMEA 10-MEAM 2-EAMM 5-AMME 8-MEMA 11-MAEM 3-EMAM 6-AMEM 9-MMAE 12-MAME I found that there were twelve arrangements for the name 'EMMA'. I then investigated different arrangements of letters in the name 'LUCY', using the same method: 1-LUCY 7-UCYL 13-CYLU 19-YCLU 2-LUYC 8-UCLY 14-CYUL 20-YCUL 3-LCUY 9-UYLC 15-CUYL 21-YUCL 4-LCYU 10-UYCL 16-CULY 22-YULC 5-LYCU 11-ULCY 17-CLYU 23-YLCU 6-LYUC 12-ULYC 18-CLUY 24-YLUC I found that there were many more arrangements in the name 'LUCY' than in the name 'EMMA'. This is because the name 'EMMA' contains repeated letters and the name 'LUCY' does not. I then investigated how many different arrangements of letters there were in other names of different lengths. The lengths of names that I used were 2, 3, and 4 letter names. I recorded my results in a table. Table showing the number of different arrangements of letters possible in words of different lengths with no repeated letters Number of letters Number of different arrangements 2 2 3 6 4 24 After recording the results in a table, I found that there was a pattern in my results. The number of arrangements of letters in a 3-letter name without repeated letters is 6, and the number
GCSE Maths Project – “Emma’s Dilemma”
Murray Goodwin GCSE Maths Project - "Emma's Dilemma" 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. Summary of Parts 1 and 2: EMMA and LUCY both have the same number of letters in their names, however LUCY has twice as many different letter combinations. This is because EMMA's name has a repeating letter (in this case the letter "M" is repeated) whereas the letters in LUCY's name are all different. Where a name has four letters (A1, A2, B, and C), the possible arrangements can be written as follows: A1A2BC A2A1BC BA2A1C CA2BA1 A1A2CB A2A1CB BA2CA1 CA2A1B A1BA2C A2BA1C BA1A2C CBA2A1 A1BCA2 A2BCA1 BA1CA2 CBA1A2 A1CA2B A2CA1B BCA1A2 CA1A2B A1CBA2 A2CBA1 BCA2A1 CA1BA2 This gives a total of 24 possible combinations. However, if we take the letters again, and this time remove the subscript numbers from the letter A, we can make the following
In my investigation I am going to investigate the number of different arrangements of letters in a word.
In my investigation I am going to investigate the number of different arrangements of letters in a word. e.g Tim Is one arrangement Mit Is another First I am going to investigate how many different arrangements in the name LUCY, which has no letters the same. LUCY LUYC LYCU LCYU LYUC LCUY ULCY UYLC UCLY ULYC UCYL There are 4 different letters and there are 24 different arrangements. SAM SMA MSA MAS ASM AMS There are 3 different letters in this name and 6 different arrangements. JO OJ There are 2 different letters in this name and there are 2 different arrangements. Table of Results Number of Letters Number of Different Arrangements 2 2 3 6 4 24 5 20 6 720 7 5040 From the table of results I have found out that a 2 letter word has 2 arrangements, and a 3 letter word has 6. Taking for example a 3 letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements. In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4!, which is called 4 factorial which is the same as 4 x 3 x 2. So, by using factorial (!) I can predict that there will be 40320 different arrangements for an 8 letter word. The formula for this is: n! = a Where n = the number of letters in the word and a = the number of
In this project I had to find out the number of different ways you could arrange any letter words with no repeats, then with one repeat and then find a formula for any number of repeats.
Kate Parkin 10HB GCSE Coursework Mathematics Emma's Dilemma Introduction In this project I had to find out the number of different ways you could arrange any letter words with no repeats, then with one repeat and then find a formula for any number of repeats. All Different Letters First I decided to do a one letter word: I There of course was only one way of arranging it! Then I did a two-letter word: Me Em There were two ways of arranging that. Next I did a three letter word You You Uoy Uyo Oyu Ouy After that I decided to do the name Lucy. I found these ways of arranging it: Lucy Lcuy Lcyu Luyc Lyuc Lycu Cuyl Cyul Cylu Culy Clyu Cluy Ulcy Ucly Uycl Uylc Ucyl Ulyc Ycul Yclu Yluc Ylcu Yucl Yulc I thought this answer was 24 because there were 4 different letters and 6 different ways to arrange it with the same first letter and so 6x4 is 24 The next name I did was Chloe. I came up with 120 ways of arranging it. For each different first letter there were 24 different ways of arranging the other letters, as ; Chloe Chleo Choel Cheol Cheol Chelo Cloeh Clohe Cleho Cleoh Clhoe Clheo Coehl Coelh Colhe Coleh Cohel Cohle Ceolh Ceohl Cehlo Cehol Celho Celoh So there would be a total of 24 x 5 = 120 ways of arranging the letters. The Formula Next I wrote out a table with my results in it: All Different Letters Ways 1 2 2 3 6 4
Emma’s Dilemma.
Emma's Dilemma Part 1 Introduction In this investigation, I am going to find out the different arrangements of the letters of the name LUCY. Here are the possibilities for the name LUCY LUCY UYCL YCLU CULY LCYU UCLY YLCU CYLU LYCU ULUC YULC CLYC LCUY ULUY YLUC CUYL LUYC UCLY YCUL CYUL LYUC UYLC YULC CLUY First of all I took the letter L from the name LUCY then wrote the second letter being U and after the two other possibilities being C and Y. Then after I used the same letter L, wrote the second letter down being C and wrote the two possibilities again being U and Y. I took another letter which I used that to be the second letter, keeping the first letter the same in till I have found the rest of the possibilities in the letter. Wrote down the 2 other possibilities. There were six possibilities for the word beginning with L. I repeated the process with the letters U, Y and C. I found out that there were 6 possibilities for each of these letters and then I multiplied 6 by 4 (the number of letters) which gave me 24 possibilities for the word LUCY. Part 2 Introduction In this investigation, I am going to find out the different arrangements of the letters of the name EMMA Here are the possibilities for the name EMMA EMMA EMAM EAMM MEMA MEAM MMEA MMAE MAEM MAME AMME AMEM AEMM First of all I took the letter M from the name EMMA then wrote the
Emma’s Dilemma
EMMA'S DILEMMA Aim To investigate the patterns caused by the permutations of letters in words of different lengths and to investigate the possibility of predicting the number of permutations. To discover a formula that can be applied to all words. Question 1. 2 different permutations can be made by the name EMMA: . EMMA 2. EMAM 3. EAMM 4. MMEA 5. MMAE 6. MEMA 7. MEAM 8. MAEM 9. MAME 0. AMEM 1. AMME 2. AEMM Question 2. 24 different permutations can be made by the name LUCY: . LUCY 2. LUYC 3. LYUC 4. LYCU 5. LCYU 6. LCUY 7. CULY 8. CUYL 9. CYUL 0. CYLU 1. CLUY 2. CLYU 3. UYCL 4. UYLC 5. ULYC 6. ULCY 7. UCYL 8. UCLY 9. YLCU 20. YLUC 21. YULC 22. YUCL 23. YCUL 24. YCLU This is double the amount of permutations made by EMMA Question 3 CLIVE . CLIVE 2. CLIEV 3. CLVIE 4. CLVEI 5. CLEVI 6. CLEIV 7. CEIVL 8. CEILV 9. CEVLI 0. CEVIL 1. CELIV 2. CELVI 3. CILVE 4. CILEV 5. CIVLE 6. CIVEL 7. CIEVL 8. CIELV 9. CVLEI 20. CVEIL 21. CVIEL 22. CVILE 23. CVLEI 24. CVLIE 25. LCEVI 26. LCEIV 27. LCVEI 28. LCVIE 29. LCIVE 30. LCIEV 31. LIEVC 32. LIEVC 33. LIVCE 34. LIVEC 35. LICVE 36. LICEV 37. LICIV 38. LECVI 39. LEICV 40. LEIVC 41. LEVCI 42. LEVIC 43. LVEIC 44. LVECI 45. LVICE 46. LVIEC 47. LVCIE 48. LVCEI 49. IVCEL 50. IVCLE 51. IVECL 52. IVELC 53. IVLCE 54. IVLEC 55. ICLEV 56. ICLVE 57.
Emma’s Dilemma.
Emma's Dilemma. Introduction The aim of this coursework is to investigate into the problem of different arrangements of a set of letters. By using the given guidelines and by widening the breadth of the operation I aim to find a rule or formulae that can represent and explain the situation. The basic parameters of the project are as follows: "Emma is playing with arrangements of the letters of her name. One arrangement is EMMA, another is MEAM, and another is AEMM. Investigate the number of different arrangements of the letters of Emma's name." This is the first task, which is then followed up by: "Emma has a friend named Lucy. Investigate the number of different arrangements of Lucy's name." After these two tasks I can then lead the investigation in my own direction: "Choose some different names. Investigate the number of different arrangements of the letters of the names you have chosen." An extension to the investigation is as follows: "A number of Xs and a number of Ys are written in a row such as 'XX . . . XXYY . . . Y . .' Investigate the number of different arrangements of the letters." The last section will allow me even more freedom to change variables to find more complex solutions covering a greater variety of situations. Finding The mathematical term for this type of function where a set of values is rearranged to give the maximum number of
Emma’s dilemma
Emma's dilemma Every name has many ways of arranging the letters e.g. BOB, OBB, BBO. I am trying to find a pattern and formula for finding out how many combinations there for a name. I tried different names with a different number of letters in them, I also tried the names with repeated letters in them. There are several methods of finding out the combinations of a name, I have used 3 different methods. Method 1 Listing Two lettered names ED DE 2 ways Three lettered names As you can see below a three lettered name with 1 repetion in it has half the number of combinations. TED TDE ETD EDT DTE DET 6 ways BOB BBO OBB 3 ways Four lettered names I have noticed that with four lettered names that with 1 letter repeated there are half the number of combinations and halved again if there are two pairs of letters LUCY EMMA LCUY EAMM LYCU EMAM LCYU AEMM LYUC AMEM LUYC AMME ULCY MMAE ULYC MMEA ULYC MEMA UYLC MAME UCLY MEAM UCYL MAEM CLUY 12 ways CLYU CYUL LULU CYLU LUUL CUYL UULL CULY ULUL YLUC ULLU YLCU LLUU YCLU 6 ways YCUL YUCL YULC 24 WAYS Since five lettered names have a far