Emma's Dilemma
Amy Cotter 11u Maths Coursework Investigation: "Emma's Dilemma" Emma's Dilemma I am going to investigate the number of different combinations of various groups of letters. LUCY is a 4 letter word with letters all different. This is the number of combination there can be for LUCY: LUCY CLUY There are 24 combinations for a 4 letter LUYC CLYU word that include letters that are all LCYU CYLU different. LCUY CYUL LYUC CULY LYCU CUYL ULCY YLUC ULYC YLCU UYLC YUCL UYCL YULC UCLY YCLU UCYL YCUL EMMA is a 4 letter word with 2 letters the same and 2 different. This is the number of combination there can be for EMMA: EMMA AEMM There are 12 combinations for a 4 letter EMAM AMME word that includes 2 letters the same and 2 EAMM MMAE different. MEMA MMEA MEAM MAEM AMEM MAME After investigating to number of combinations there are for the words LUCY and EMMA, I am going to start with the simplest case; which is 1 letter: A There is 1 combination for 1 letter Next I am going to increase the number of letters by one each time
Emma's Dilemma
27th June 2001 Tom Pountain 10A Emma's Dilemma I investigated the number of different arrangements of four letters with no repetitions. )ABCD 2)ABDC 3)ADBC 4)ADCB 5)ACBD 6)ACDB 7)BACD 8)BCAD 9)BDCA 0)BADC 1)BDAC 2)CDBA 3)CBAD 5)CDAB 6)CADB 7)CBDA 8)DABC 9)DBCA 20)DCAB 21)DACB 22)DBAC 23)DCBA 24)DABC 25)BCDA I have found 24 different arrangements of these letters and this result is confirmed in the tree diagram. Secondly, I have investigated the number of different arrangements of four letters with one letter repeated twice. )ACBB 2)ABCB 3)BABC 4)BACB 5)BBAC 6)BBCA 7)BCAB 8)BCBA 9)CBBA 0)CABB 1)CBAB 2)ABBC I have found 12 different arrangements of the letters and this result is confirmed in the tree diagram. From these two investigations, I have worked out a method that can be used for further work: Firstly, with ABCD you rotate the last two letters, then you get ABDC. Then, with ABCD you must then rotate the last three letters and try the possibilities of ADBC, ADCB, ACDB, ACBD. Because the letter 'B' has been the first number of last three letters before, we don't do it again. We have list all arrangements with A first, so now we do B secondly: BACD, and we do same thing to it, it will like this: BACD=BADC, BADC=BDCA...BDAC...BCAD...BCDA We
Emma's Dilemma
EMMA is investigating the amount of different arrangements of letters in her name; she does the same with her friend LUCY. LUCY has twice as many arrangements as EMMA, they are curious as to why this is and decide to investigate other names and find reasons for their answers. EMMA - emma, eamm, emam, aemm, amme, amem, meam, maem, mame, mema, mmea, mmea, LUCY - lucy, luyc, lycu, lyuc, lcyu, lcuy, ulcy, ulyc, uylc, uycl, ucly, ucyl, cluy, clyu, culy, cuyl, cyul, cylu, yluc, ylcu, yulc, yucl, yclu, ycul, Emma has 12 combinations and Lucy has 24. 1/2 of 24 = 12 so Emma has half the amount that Lucy has; this may be because Emma has 2 letters the same. I will be investigating whther this happens to other names as well. I will Investigate - * The amount of combinations for names with 2-10 letters * What happens when those names have 2 letters the same * What happens when they have 2,3,4,5 etc. letters the same * Whether 3 letters the same means 1/3 of the combinations it would have if no letters were the same * Whether 4 letters the same means 1/4 of the combinations it would have if no letters were the same (and 5, 6 ,7 etc.) * Whether there are any patterns or rules to follow when estimating amounts of combinations * What happens when words have more than 1 letter twice (e.g. LIANNA) 2 letter - 0 same = 2 JO,
Emma's dilemma.
I am investigating the number of different combinations of letters for the name Emma. * EMMA 12 combinations * EMAM * EAMM * MEMA * MAME * MMAE * MMEA * MEAM * MAEM * AMME * AMEM * AEMM The name Emma has 12 different combinations. As the name Emma has two M's in it I will investigate the number of different combinations for the name Lucy. * LUCY 24 combinations * LUYC * LCYU * LCUY * LYCU * LYUC * UCLY * UCYL * ULCY * ULYC * UYLC * UYCL * CUYL * CULY * CLUY * CLYU * CYUL * CYLU * YCLU * YCUL * YULC * YUCL * YLCU * YLUC The name Lucy has 24 combinations this is twice as many combinations as Emma this is because EMMA has two letters the same. If you were to colour code the Ms in EMMA then you may have 24 combinations. I am going to investigate this. * EMMA * EMMA* * EMAM * EMAM* * EAMM * EAMM* * MEMA * MEMA* * MAME * MAME* * MMAE * MMAE* * MMEA * MMEA* * MEAM * MEAM* * MAEM * MAEM* * AMME * AMME* * AMEM * AMEM* * AEMM * AEMM* All of the combinations with a * next to them are cancelled out because if they were not colour coded then they would look exactly the same as the combination above it. That is why there are twice as many LUCYs to EMMAs. I am now going to investigate some other names that I have chosen. 2 letters * Jo 2 combinations * Oj 3 letters * Sam 6 combinations * Sma * Ams * Asm * Mas
Emma's Dilemma
Emma's Dilemma Kevin Etuk 28/3/02 Here are my Different arrangements for the letters in the name Emma: Emma , emam, eamm, mmae, mmea, meam, amem, aemm, Mema, mame, maem, amme, amem, aemm, There are 12 different variations you can make from the letters in the name Emma. Lucy Here are my Different arrangements for the letters in the name Lucy: Lucy, ucyl, cylu, ycul, luyc, ucly, cyul, yclu, lcuy, ulcy, culy, yulc, lcyu, ulyc, cuyl, yucl,lyuc, uycl, clyu, ylcu,lycu, uylc, cluy, yluc I have found 24 Variations you can make with the name Lucy. If you take the one letter from Emma you can make 3 different Arrangements. If you Times 3 by 4 you get 12 which is the total number of variations. I I'm going to see if this is true for Lucy. Yes if you take a letter you can make 6 different Variations. Times 6 by 4 you get 24, which is the total number of variations. Rule for a 4 letter word with 4 different letters 4 X 6 = T(total number of variations) Rule for a 4 letter word with 3 different letters is 4 X 3 = T(total number of
Emma's Dilemma.
Emma's Dilemma Coursework Aim: To investigate the number of different arrangements of the letters in the name Emma and several other names I have chosen and find a formula for the names with different amounts of letters. I will also investigate the number of different arrangements of the letters X and Y and find a formula for the words with different amounts of X's and Y's. Prediction: I think that the more letters in the name the more arrangements of that name there will be. I also think that there will be more arrangements of Lucy than of Emma because all the letters in Lucy are different whilst two letters in the name Emma are the same. Method: To work out the number of different arrangement of names I will write down the first letter of the name, then systematically work out all the arrangement of the name that begin with that letter by using the branching method. The branching method looks like this Y C LUYC U C Y LUCY L U Y LCUY C Y U LUYC U C LYUC Y C U LYCU As you can see there is six different arrangement of the name Lucy that begin with L. If I was only looking for the number of arrangements of the name Lucy I could the multiply the number of arrangements that begin with L by how many letters there are in the name (this only works if all the letters in the name are different) or I could work out all the other arrangements by using the
Emma's Dilemma
Emma's Dilemma Plan I am investigating the number of different arrangements there will be in different types of names. Some names I will investigate on will have no identical letters such as LUCY. Some will have a pair of identical letters such as EMMA. Some names will have different quantities of letters such as AMMIE, JOE and ANNE. Firstly, I'll produce a method which will help me figure out the different arrangements in the name EMMA and LUCY without using any formulas. Using this method I'll test all my predictions. I'll create formulas for different types of names. Examples are names with no identical letters, a pair or more identical letters and names that have 2 different groups of identical letters Throughout the investigation I will devise different formulas for working out different types of names. By using one of these formulas I'll figure out the total number of arrangements of the letters XX......XXYY.......Y. To start of the investigation I have to find the different arrangements of letters in the name EMMA and LUCY without the use of any types of formulas. Therefore, I have developed my own method. Method . Write the name in its original format. In this case it will be EMMA or LUCY. 2. Start with the first letter in the original format (E in EMMA and L in LUCY) and rearrange the remaining letters in a random order in front of the first letter. 3.
Emma's Dilemma.
Lee Hamilton Emma's Dilemma Emma is playing with the arrangements of her name, find out how many combinations there are I will write down the combinations of Emma by doing each letter at a time swapping the remainder of the letters around, so when I am further into the experiment I can work out the amount of combinations in word by looking at my result tables. 4 letters 3 different Arrangements for Emma: ) emma 2) emam 3) eamm 4) mmae 5) mmea 6) meam 7) mema 8) mame 9) maem 0) amme 1) amem 2) aemm There are 12 possibilities when there are 4 total letters and 3 different. From this you could make an hypothesis that the formula could be n (number of letters) x the amount of different letters. I used mm as 1 combination instead of making them m1 and m2. This is becuase you are finding different combinations and mm is esacially the same every time and can't be changed to look different. To explain better you can just think of colour cubes. If you have a green, red, blue and blue and u changed the two blues around it still would look the same. You also notice that there are 3 combinations for the letter a and e but 6 for m. This is because it just acts as two letter when it is at the front of the word as the other m can move around in the word. Emma's friend Lucy then decides to play with the arrangements of her name, find out how many combinations there
Emma's Dilemma
Emma's Dilemma I have been asked to find out how many ways I can arrange the EMMA. I have then been asked to work out a formula for this without writing down every combination. I will now find out how many ways I can arrange the letters EMMA. One combination is EMMA Another is EMAM EAMM MMAE MAME MEMA MAEM MEAM MMEA AEMM AMME AMEM They are 12 different combinations of EMMA. 4 letters but 3 different letters I will now work out how many combinations they are with a 4 letters all different I will try PHIL one combination is PHIL ILPH PHLI ILHP PIHL IHPL PILH IHLP PLHI IPLH PLIH IPHL HILP LIHP HIPL LIPH HPIL LHPI HPLI LHIP HLIP LPIH HLPI LPHI I have found out that they are 24 different combinations. 4 letters all of them different. I have worked that this is double to EMMA. I have also worked out that with every 2 letters beginning they are 12 combinations so with every one letter beginning they are 6 different combinations. I will now try 5 letters and find out how many different combinations they are. I predict that they will be 120 different combinations. I predict this because they is 24 combinations for PHIL so if you added another letter you could put the new letter at the beginning of each combination. I will try and find out how many combinations they are with PHILM. MPHIL MILPH MPHLI MILHP MPIHL MIHPL
Emma's Dilemma
GCSE COURSEWORK Matthew Kilgour 25/8/01 Arrangements for Emma: 4 letters, 2 the same. .emma 2.emam 3.eamm 4.mmae 5.mmea 6.meam 7.mema 8.mame 9.maem 10.amme 11.amem 12.aemm In a word/name with 4 letters and none the same, then there are 24 possible arrangements. There are 12 possibilities; note that there are 4 total letters and 3 different. Emma has four letters, three of which are different. This means that it only has 12 possible combinations. I had noticed that 4 (number of total letters) multiplied by 3 (the number of different letters) equals 12. What if all the letters were different like in Lucy? Arrangements for Lucy: 4 letters, all different, none the same. As I stated before, in a word with 4 letters and none the same, there are 24 possible combinations, double the number of combinations for Emma, which has 4 letters and 3 the same. .lucy 2.luyc 3.lcyu 4.lycu 5.lcyu 6.lyuc 7.ulcy 8.ucly. 9.uylc 10. ulyc 11.uycl 12. ucyl 3.yucl 14.ycul 15.yluc 16. ylcu 17.yclu 18.yulc 9.cluy 20.clyu 21.cylu 22.cyul 23. Culy 24.cuyl noticed that with Lucy, there are six combinations starting with each letter. Eg. 6 with L, 6 with U etc. I had also noticed that 6 x 4 (the number of letters) is 24, the total number of possibilities. How many combinations would there be