Emma’s Dilemma
A.M.D.G June 2001 Emma's Dilemma I firstly found the different number of arrangements for Emma's name. . EMMA 2. EMAM 3. EAMM 4. AEMM 5. AMEM 6. AMME 7. MAME 8. MAEM 9. MEAM 0. MEMA 1. MMAE I then found the different number of arrangements for Lucy's name. . LUCY 2. LUYC 3. LYUC 4. LYCU 5. LCUY 6. LCYU 7. ULCY 8. ULYC 9. UYLC 0. UYCL 1. UCYL 2. UCLY 3. CULY 4. CUYL 5. CYUL 6. CYLU 7. CLYU 8. CLUY 9. YLUC 20. YLCU 21. YCLU 22. YCUL 23. YUCL 24. YULC The method I used to investigate the number of different arrangements of Emma's name was by keeping the first two letters the same and rearranging the last two letters. I kept on doing this until I had the maximum number of arrangements, which was 12. I repeated this procedure with Lucy's name and found 24 arrangements. Next I found the arrangements of some different names to see if I could spot a pattern. . TOM 2. TMO 3. MTO 4. MOT 5. OMT 6. OTM My strategy for finding the number of arrangements for the name Tom, was by this time, keeping only the first letter the same and rearranging the last two letters. 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
Emma’s dilemma.
Emma's dilemma Edexcel Maths Course Work CANDIDATE NAME: Akbar Miah SCHOOL NAME : London Islamic School/Madrassah Emma's Dilemma 5. In this piece of work I will be investigating all the different arrangement of words, and try to find formulas that will enable us to calculate how many different ways you can arrange various lettered words. Example the word car is one way of arranging it and arc is another way of arranging it. This coursework is divided into three sections I will try to find the different arrangements of the letters of the following names: Part 1: Lucy is a four-lettered word, all the letters are different. lucy ucly cluy yucl lcyu ucyl cuyl ycul lycu uylc culu yluc lcuy ulyc cluy yccl lyuc ulcy culy yclu lyuc uycl cluy ylcu With Lucy's name you can arrange it 24 different ways. I could also calculate all the different arrangements of this word by only working out the number of times Lucy's name can be arranged when it begins with the letter l, u, c or y, for example the word Lucy can be arranged 6 times when it begins with l, so because there are four letters you can multiply 6 by four, which will give you the number of times you can arrange the word Lucy, which is 24, and as we can see from above this theory is correct. Part 2: Emma: Emma is also a four-lettered
Emma’s Dilemma
Emma's Dilemma I have been given the task of finding different combinations for different names with different amount of letters. I am to find the right formula that will work for all names. I have been given two names. "Lucy" and "Emma" and I am to find a formula that will hopefully help me to work out all the combinations for any word that is given to me. Firstly I will take the name Lucy because I have notice that in the name Emma there is a repeated letter and so I will leave this for later. Lucy is a four letter word and it would make sense to firstly start with a two letter word and work my way up to four so that I may be able see a trend or pattern. I will take the name "Jo": - jo Oj As you can see there are only 2 different combinations for the name Jo. The next name I will take is "Sam": - sam ams msa Sma asm mas The name Sam has 6 different combinations. Now I have come to a four letter word I can work out the combinations for "Lucy": - lucy ulcy cylu yluc Luyc ulyc cyul ylcu lcuy uylc cluy ycul lcyu uycl clyu yclu lycu ucyl cuyl yucl lyuc ucly culy yulc As you can see the name Lucy has 24 different combinations. I will now put the results of these names into a table and then see if I can come up with a formula that will help me work out the number of combinations for any number of letters that are all different. Table of
Emma’s Dilemma.
Emma's Dilemma This task involves playing with different arrangements of different names. The names will vary in size and with letters being the same or not the same. Different arrangements for the name EMMA are: ) EMMA 4) AEMM 7) MAME 10) MEAM 2) EMAM 5) AMEM 8) MMAE 11) MAEM 3) EAMM 6) AMME 9) MEMA 12) MMEA Different arrangements for the name LUCY are: ) LUCY 7) ULCY 13) CULY 19) YCLU 2) LUYC 8) ULYC 14) CUYL 20) YCUL 3) LYUC 9) UYLC 15) CYLU 21) YULC 4) LYCU 10) UYCL 16) CYUL 22) YUCL 5) LCUY 11) UCYL 17) CLUY 23) YLUC 6) LCYU 12) UCLY 18) CLYU 24) YLCU I am going to find out if there is a relationship between number of letters and number of arrangements when all the letters in the combination are different. L has only one arrangement which is L. LU has two arrangements: ) LU 2) UL LUC has six arrangements: ) LUC 3) ULC 5) CLU 2) LCU 4) UCL 6) CUL LUCY, as we know has 24 different arrangements. Here is a table showing the number of letters and the number of arrangements when all the letters are different: Number of Letters Number of Arrangements 2 2 3 6 4 24 The number of arrangements is equal to the number of letters multiplied by the previous number of letters. For example, 4 x 3 x 2 x 1 = 24 Therefore LUCEY must have 120 different combinations because it has 5 letters and 5 x 4 x 3 x 2 x 1 = 120 LUCEY 1) LUCEY 7) LEUCY 13)
Emma’s Dilemma
John Palfrey, 5UL Emma's Dilemma ) Rearrangement Of EMMA If we represent the letters in Emma's name as numbers, we get an arrangement like this: E M M A 1 2 2 3 We can then use this system to find all of the other potential arrangements: 1 2 2 3 1 2 3 2 1 3 2 2 2 1 2 3 2 1 3 2 2 2 1 3 2 2 3 1 2 3 1 2 2 3 2 1 3 1 2 2 3 2 1 2 3 2 2 1 As you can see, there are 12 combinations. 2) Rearrangement Of LUCY If we represent the letters in Lucy's name as numbers, we get an arrangement like this: L U C Y 1 2 3 4 We can then use this system to find all of the other potential arrangements: 1 2 3 4 2 1 3 4 1 2 4 3 2 1 4 3 1 3 2 4 2 3 1 4 1 3 4 2 2 3 4 1 1 4 2 3 2 4 1 3 1 4 3 2 2 4 3 1 3 1 2 4 4 1 2 3 3 1 4 2 4 1 3 2 3 2 1 4 4 2 1 3 3 2 4 1 4 2 3 1 3 4 1 2 4 3 1 2 3 4 2 1 4 3 2 1 As you can see, there are only 24 potential arrangements; twice as many as when there is one repetition. 3) Other Rearrangements If we use the name MAY to investigate the number of potential arrangements when there are only three factors, we get these arrangements: M A Y 1 2 3 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 We only get six arrangements, a quarter of the amount when we use four factors. If we use the word MOO to investigate the number of potential arrangements when there are only three factors, we get these arrangements: M O O 1 2 2 1 2 2 2 1 2 2 2 1 We only
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 . Investigate the number of different permutations of the letters of the name Emma. I am trying to find the maximum number of possible permutations of the name EMMA. This name has four letters but only three variable letters E, M and A. Permutations: EMMA MMAE AEMM EMAM MMEA AMEM EAMM MAME AMME MAEM MEMA MEAM This shows us that there are twelve possible permutations of the letters of the name EMMA. Emma has a friend called Lucy. 2. Investigate the number of different permutations of the letters of the name Lucy. I am now trying to find the maximum number of possible permutations of the name LUCY. This name has four letters any four variable letters L, U, C, and Y. Permutations: LUCY ULCY CLYU YLUC LUYC ULYC CLUY YLCU LCYU UYCL CULY YCUL LCUY UYLC CUYL YCLU LYCU UCLY CYUL YULC LYUC UCYL CYLU YUCL This shows us that there are twenty-four possible permutations of the letters of the name LUCY. Chose some different combinations of letter. 3. Investigate the number of different permutations of the letters that you have chosen. I am now going to try to find a pattern in the permutations of combinations of different letters. a) Firstly with one letter. * 1 A There is one permutation with one letter. b) With two letters. * 2 AB BA There are two permutations with two different letters. c) With three letters. * 3 ABC BAC
Emma's Dilemma
Mathematics GCSE EMMA'S DILEMMA . No. of arrangements Arrangements EMMA 2 MEAM 3 AEMM 4 MMEA 5 MMAE 6 AMEM 7 EMAM 8 AEMM 9 MAEM 0 AMME 1 MAME 2 MEMA 2. No. of arrangements Arrangements LUCY 2 LYCU 3 LYUC 4 LUYC 5 LCYU 6 LCUY 7 UCYL 8 UYCL 9 ULCY 0 ULYC 1 UYLC 2 UCLY 3 CYLU 4 CYUL 5 CLUY 6 CLYU 7 CULY 8 CUYL 9 YLUC 20 YLCU 21 YUCL 22 YULC 23 YCLU 24 YCUL 3. No. of arrangements No. of letters Position(n) A 2 AB 2 6 ABC 3 24 ABCD 4 20 ABCDE 5 720 ABCDEF 6 5040 ABCDEFG 7 40320 ABCDEFGH 8 362880 ABCDEFGHI 9 3628800 ABCDEFGHIJ 0 I have completed the table by using a method I found after finding a pattern in the first few results. 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. of arrangements) (all different) 5 letter word 1x2x3x4x5=120 (all different) 6 letter word 1x2x3x4x5x6=720 (all different) I then finished off the table up to a 10-letter word (all different letters). Next I decided to do a further
Emma's Dilemma
GCSE Mathematics Coursework Emma's Dilemma In my investigation I am going to investigate the number of different arrangements of letters for names and words and try to find a formula that can be used to predict this. For example: TOM is one arrangement and OTM is another arrangement First, I am going to investigate the number of different arrangements of letters for the name LUCY (a 4-letter name, where all the letters are different). LUCY ULCY CLUY YLUC LUYC ULYC CLYU YLCU LCUY UCLY CULY YULC LCYU UCYL CUYL YUCL LYUC UYLC CYLU YCLU LYCU UYCL CYUL YCUL There are 4 different letters and 24 different arrangements. Once I have investigated the number of different arrangements for one 4-letter name/word where all the letters are different, I do not need to try any more. If I tried the name DAVE for example, there would still be 24 different arrangements. I could substitute the L in LUCY for the D in DAVE, the U for A, the C for V, and the Y for E; and would therefore end up with the same result. The same is true for names/words with 3 letters or 5 letters, etc. As long as the number of letters and the number of different letters are the same, the number of different arrangements will be the same. Now I will investigate a 3-letter name where all the letters are different. SAM ASM MSA SMA AMS MAS There are 6 different arrangements. Now I
Emma's Dilemma
Matthew Shaw/10Ds/Mrs Finn 17/07/02 Mathematics GCSE Emma's Dilemma On this piece of mathematics coursework called Emma's Dilema we are going to look at all the different ways that we can arrange names. On this piece of coursework I am trying to achieve to find all the different ways I can arrange names and also I want to find out a good enough formula so that i can find the different arrangements more easily. Here are the different arrangements of writing EMMA's name: EMMA EAMM EMAM MEAM MAEM MMEA MMAE MEMA MAME AEMM AMEM AMME Here are the different arrangements of LUCY's name: LUCY LCUY LYCU LYUC LCYU CLUY CULY CYLU CLYU CUYL CYUL ULCY ULYC UCYL UCLY UYLC UYCL YLUC YLCU YULC YUCL YCLU YCUL LUYC In my investigation that I undertook from my results I found out that there were 24 different ways of writing LUCY, and only 12 ways of writing EMMA. This is because there are 2 M's, so therefore this means that there are 3 letters to do your combination of EMMA's name, and in LUCY's name there are 4 so therefore this cuts down the number of times you can arrange the letters, even thou there are four letters in each name. In my next investigation I am going to choose different names with 2,3,4 and 5 letters in them. These are the names I will investigate JO, SAM, LUCY, SIMON. 2 letters = JO =2 5