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# GCSE: Emma's Dilemma

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1. ## Emma's Dilemma

gives 24. This means that all names with 4 different letters will have a total of 24 different combinations. I realised that the total number of combinations is the number of letters multiplied by the previous number of combinations. Name: Ian Total number of letters: 3 Previous number of combinations: 2 3 X 2 = 6 Name: Lucy Total number of letters: 4 Previous number of combinations: 6 4 X 6 = 24 This means that I can work out the total number of combination by factorial notation. Factorial notation is a number multiplied by the previous consecutive numbers: E.g.

• Word count: 1491
2. ## Emma's Dilemma

Rmay Hnjo 19 Yclu Yamr Nojh 20 Ycul Yarm Nohj 21 Yucl Ymar Nhjo 22 Yulc Ymra Nhoj 23 Ylcu Yram Njoh 24 Yluc Yrma Njho Total rearrangements for words with 4 letters, none of them being the same letter = 24 1 Emma Jazz 2 Emam Jzaz 3 Eamm Jzza 4 Amme Ajzz 5 Amem Azjz 6 Aemm Azzj 7 Mmae Zzja 8 Mmea Zzaj 9 Maem Zazj 10 Mame Zajz 11 Mema Zjaz 12 Meam Zjza The total rearrangements for words with 4 letters, 2 of them being the same letter = 12 Jzzz Emmm Zjzz Memm Zzjz

• Word count: 1872
3. ## Emma&#146;s Dilemma

LOU ULO OUL LUO UOL OLU The name LEE has 3 letters, 2 the same. LEE EEL ELE The name CCC has 3 letters, all the same. CCC has no variations. 2 letters The name LI has 2 letters, both different. LI IL The name DD has 2 letters, both the same. DD has no variations. After carrying out these experiments, I have come up with the following table of results: No. of letters All different 2 the same 3 the same 4 the same 2 2 1 3 6 3 1 4 24 12 4 1 From this table I see a pattern.

• Word count: 866
4. ## Emma&#146;s Dilemma

However this is only the case with 3 letter words. JO OJ There are 2 different letters in this name and there are 2 different arrangements. From these investigations I worked out a method: Step1: 1234---Do the last two numbers first then you get 1243. 1243---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.

• Word count: 1196
5. ## Emma&#146;s Dilemma

UCYL CULY CUYL CYLU CYUL CLUY CLYU YCUL YCLU YLUC YLCU YUCL YULC I am now going to put this into a table of results: Number of letters in a word Number of arrangemens 1 1 2 2 3 6 4 24 I have found a formula for this. I found that if you factorise the number of letters in a word, you get the number of arrangements. If you times 1 x 1, you get 1 arrangement. If you time 1 x 2, you get 2 arrangement.

• Word count: 1298
6. ## Emma&#146;s Dilemma

LUCY The name Lucy has no double letters but still has four letters. I will investigate Lucy to see if there are a different number of arrangements. In the name Lucy, there are six different arrangements for each letter (L, U, C, Y,) this is different from EMMA as In the name EMMA, there was only three options for each letter apart from M as M was a double letter. I would expect LUCY to have more options as the first letter, has three different letters that can be rearranged after it.

• Word count: 1288
7. ## The relationships between the number of different spacers in an arrangement of square tiles and the dimensions of the tiles in the same arrangement.

Therefore the rule for the L shaped spacers will be n = 4. For the + spacers the rule will be n = (n-1) . For the T Spacers the rule will be n = 4n - 4. I will now test these rules. I have now drawn the 10x10 arrangement and have written the actual properties as read from my sketch in this table: Pattern number Number of Squares + Spacers T Spacers L Spacers 10 100 81 36 4 From the results gathered by actually drawing the arrangement I have deduced that the rules that I obtained from

• Word count: 966
8. ## Emma's Dilemma

For a four-letter word (all different letters) there are 24 arrangements as I found in Question 2 (LUCY). There are six combinations beginning with each letter, with this knowledge I think that a five-letter word (all different letters) would have 120 arrangements. To prove this answer/back it up I have found another pattern in the table. 4 letter word 1x2x3x4=24 (No.

• Word count: 587
9. ## Emma's Dilemma.

Investigation 1 I am using these names because none of the letters in them are doubled. The method which I will use to re-arrange the letters is a clockwise rotation. 2 letter name ~ AL AL LA 3 letter name ~ SAM MSA AMS MAS ASM SAM SMA 4 letter name ~ LUCY LUCY UCLY CYLU YUCL LUYC UCYL CYUL YULC LYCU UYLC CLYU YLCU LYUC UYCL CLUY YLUC LCYU ULCY CULY YCUL LCUY ULYC CUYL YCLU 5 letter name ~ KATIE KATIE ATKIE TIEAK IEKAT EKATI KAITE ATKEI TIEKA IEKTA EKAIT KATEI ATEIK TIKAE IETAK EKITA KAETI ATEKI TIKEA IETKA EKIAT KAIET ATIKE TIAKE IEAKT EKTIA KTAIE ATIEK TEIAK IEATK EKTAI KTIAE AEKIT TEAIK IKEAT

• Word count: 1752
10. ## Emma's Dilemma

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 letters and that there were 6 ways begging with one of the letter I predict that there will be 120 arrangements for lucyq, 24 for a, 24 for b, 24 for c; ect.

• Word count: 1011
11. ## Emma&#146;s Dilemma

CEIVL 8. CEILV 9. CEVLI 10. CEVIL 11. CELIV 12. CELVI 13. CILVE 14. CILEV 15. CIVLE 16. CIVEL 17. CIEVL 18. CIELV 19. 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. ICVLE 58. ICVEL 59. ICEVL 60. ICELV 61. ILEVC 62. ILECV 63. ILVCE 64. ILVEC 65. ILCEV 66.

• Word count: 1073
12. ## Emma's Dilemma

I used the same principle for Lucy's name. Here are some other 4-letter names where one has 2 letters the same, and the other one all letters are different. I am doing this to test that the results are the same as Emma and Lucy. Arrangements for Elly: Elly Elyl Eyll Lely Leyl Lyel Lyle Lley Llye Yell Ylel ylle There are 12 arrangements for Elly's name. Arrangements for Mark: Mark Makr Mkar Mkra Amrk Amkr Armk Arkm Akrm Akmr Amkr Amrk Rmak Rmka Rkma Rkam Ramk Rakm Kmar Kmra Krma Kram Karm kamr There are 24 different arrangements for Mark's name.

• Word count: 878
13. ## Emma&#146;s Dilemma.

However, there are two Ms and I must be aware of this repeating letter. I have found that if I replace the letters in the name and assign each to its own number then I can arrange combinations of the name numerically, thus avoiding errors from overlooking any combinations or from duplicating any permutations. Therefore: E = 1 M = 2 A = 3 I have numbered both the Ms as 2, which indicates that they are indistinguishable. I can start with the lowest value combination of 1223 and go through up to 3221 (that is, EMMA through to AMME respectively): Combination # Combination Letter Representation 1 1 2 2 3 E M M

• Word count: 5964
14. ## Emma&#146;s dilemma

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 greater number of combinations, I will use another way of listing them, I have noticed that CLARK has a four lettered word after the C with no letters repeated, so CLARK beginning with a C must have 24 combinations, therefore must be the same with the word beginning with L, A, R AND K.

• Word count: 922
15. ## Emma&#146;s Dilemma

To show that this rule works, I checked it with a three-letter name. To begin with I took all the arrangements starting with the same letter for the three-letter name of Amy. These were AMY AYM If my prediction was right then No of letters in word (L) x arrangements for one letter = total number of arrangements (A) 3 ( A + M + Y) x 2 (AMY + AYM) = 6 I checked this by finding out the rest of the arrangements AMY + MAY YMA AYM MYA YAM The total number of arrangements was 6 showing that the prediction and the rule also works with the three letter word Amy.

• Word count: 3404
16. ## Emma&#146;s Dilemma

njsao njosa njoas njaso njaos nsjoa nsjao nsoja nsoaj nsajo nsaoj nojsa nojas nosja nosaj noajs noasj najso najos nasjo nasoj naojs naosj 120 Why does 'n!' give the answer? To find out why the formula n! gives the answer we need to divide up the possibilities of the word into the different possibilities of each letter. This is best demonstrated in a tree diagram In the name Jack, there are 4 letters; this means that for the first letter there are 4 different possibilities. The remaining part of the word then has 3 different possibilities (the 3 other letters).

• Word count: 1273
17. ## Emma&#146;s Dilemma

If we take, for example, a three-letter word, I have worked out that if we do 3 (which is the length of the word) ? 2 ?1= 6 which is the number of different arrangements. In a four letter word, to work out the amount of different arrangements, you can do 4 ? 3 ? 2 ? 1= 24. As you can see from the above table that this is correct. You can also do this process on a calculator, by using the factorial button, which looks like !

• Word count: 1378
18. ## Emma&#146;s Dilemma

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 combinations per starting letter) = 24 (total combinations). Now that I have investigated the number of possible rearrangements involving four letters with no similar letters and one similar letter, I should investigate the number of possibilities with tow and three similar letters.

• Word count: 1547
19. ## Emma&#146;s Dilemma

PROOF J A M I E * JAMIE JAMEI JAEIM JAEMI JAIME JAIEM JMAIE JMAEI JMEIA JMEAI JMIEA JMIAE JIMAE JIMEA JIEMA JIEAM JIAEM JIAME JEAIM JEAMI JEMAI JEMIA JEIMA JEIAM So there are 5 starting letters and so therefore 5 x 24* = 120 arrangements. Another formula for this is: n! = a Where n = the number of letters in the word and a = the number of different arrangements. The reason for this formula is that when you have a 5-letter word there are 5 spaces to out the first letter in _ _ _ _

• Word count: 1710
20. ## Emma&#146;s Dilemma

I kept on doing this until I had found the maximum number of arrangements, which was 6. In order to get an orderly pattern, I then decided to find the arrangements of a name with two different letters and one letter. 1. J One letter gives just the one arrangement. 1. JO 2. OJ Two different letters give two arrangements. I don't think I need any more arrangements of names because I think you can get a simple formula from four names.

• Word count: 1772
21. ## Emma's Dillemma - Rearranging Emma's Name in different permutations

Maybe it would be easier to see what is happening if I used larger words. 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 letters and that there were 6 combinations beginning with each of the letters in the word I predict that there will be 120 arrangements for qlucy, 24 for q, 24 for l, 24 for u; and so on.

• Word count: 951
22. ## Emma's Dillemma

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. 120 combinations will be too much so I will do a 5-letter word that has a repeated letter such as Barry Larry or Annie.

• Word count: 1561
23. ## Emma's Dilemma

I have also noticed that there are only 6 possible combinations per letter. So there are 4 letters in the word. If I multiply 6 with 4, that gives me 24 different combinations. At this stage, I am not sure that this method will work for other words with more than 5 letters, but later on in the investigation, I will try to apply this method to such words. I will now choose another word with 4 different letters in it, and I will confirm the theory that there are 24 different combinations of letters in it. 1. JOHN 2.

• Word count: 1389
24. ## Emma's Dilemma

Let's now try this with 5 letters and no repetitions: 1. ABCDE 2. ABCED 3. ABECD 4. ABEDC 5. ABDEC 6. ABDCE 7. ACBDE 8. ACBED 9. ACEBD 10. ACEDB 11. ACDEB 12. ACDBE 13. ADBCE 14. ADBEC 15. ADEBC 16. ADECB 17. ADCEB 18. ADCBE 19. AEBCD 20. AEBDC 21. AEDBC 22. AEDCB 23. AECDB 24. AECBD The arrangements of the last 4 letters with 'A' first added up to 24 again, so if we timed 24 by 5 we would get 120, and 120 is also the total of arrangements of 5 letters according to the tree diagram.

• Word count: 3071
25. ## Emma&#146;s Dilemma

> Investigating the number of arrangements in a two-letter word with 2 different letters. 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 5 120 6 720 7 5040 > From the table of results I have found out that a 2-letters word with individual letters has 2 arrangements, and a 3-letter word with individual letters has 6. > Taking for example a 3-letter word, I have worked out that if we do 3 (the length of the word)

• Word count: 1116