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# GCSE: Emma's Dilemma

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1. ## GCSE Maths questions

• Develop your confidence and skills in GCSE Maths using our free interactive questions with teacher feedback to guide you at every stage.
• Level: GCSE
• Questions: 75
2. ## Emmas Dilema

3 star(s)

Combinations with no repeats: A: 1. A AB: 1. AB 2. BA ABC: 1. ABC 2. ACB 3. BAC 4. BCA 5. CAB 6. CBA ABCD: 1. ABCD 2. ABDC 3. ACBD 4. ACDB 5. ADCB 6. ADBC 7. BACD 8. BADC 9. BCAD 10. BCDA 11. BDAC 12. BDCA 13. CABD 14. CADB 15. CBAD 16. CBDA 17. CDAB 18. CDBA 19. DABC 20. DACB 21. DBAC 22. DBCA 23. DCAB 24. DCBA Here is a table of my results: Amount of letters in combination Amount of different combinations made by letters 1 (A)

• Word count: 863
3. ## Emma's Dilema

TIM BOB TMI BBO IMT OBB ITM MTI MIT (Number of letters�2)�(Number of repeated letters) works for the 3 letter words but can't work for the 4 letter words because 6� does. No. of letters No. of repeats Multiplier of letter no. LUCY 4 0 6 EMMA 4 1 6 NOON 4 2 ? TIM 3 0 2 BOB 3 1 2 The formulae can't work for any other letter words, and the 4 letter word equation didn't work for 2 repeats in a 4 letter word.

• Word count: 1007
4. ## Emma's Dilemma

This means that to find the number of arrangements of letters in a word you use this expression: N! N! means N factorial. This means that you times N by all of the numbers below it until you get to 1. N x (N-1) x (N-2) x (N-3) x (N-4) x (N-5) .......... x 1 Using this expression I am going to work out how many different letter arrangements I think there will be for these names with different numbers of letters and then check them by writing out the arrangements.

• Word count: 1249
5. ## emmas dilema

If there are three letters like ABC then there are 6 arrangements. I organized this so that set 1 begins with A, set 2 with B and set 3 with C. Set 1 Set 2 Set 3 ABC BAC CAB ACB BCA CBA If there are four letters like LUCY then there are 24 arrangements as shown at the beginning of part 1. Results Table Number of letters Arrangements 1 1 2 2 3 6 4 24 I constructed a table to see if there were any connections between the number of letters and the different arrangements.

• Word count: 2471
6. ## Emma's dillemma

"Lucy" is a four letter word that has all the letters different in it. By working out how many permutations of the word "Lucy" I will be able to notice a pattern between words that have all the letters different in them. This will help me with my investigation as it will help me work out the pattern between words with all the letters different, and so I will be able to find out a word that has many letters in it without a problem as long as they are all different.

• Word count: 2422
7. ## Emma's Dilemma

This is because 'Emma' is a four lettered word. This was a much trickier word, as it had a repeat letter, that complicated the issue, and hence the outcome is different to 'Lucy' which is also a four lettered word. I have indicated how I have split the word up whilst arranging it. Part 3 In this part of the investigation I will be I will be working out how many possible arrangements there are for words with different lengths and different letters.

• Word count: 3184
8. ## Emma's Dilemma

and the number of combinations will still be the same. But if you were to make the double letters different I predict that there will be the same amount of arrangements as Lucy's name - 24. I will now try to prove this: EMMA MMAE MMAE AEMM EMMA MMEA MMEA AEMM EMAM MAEM MAEM AMME EMAM MAME MAME AMEM EAMM MEMA MEMA AMEM EAMM MEAM MEAM AMME Each of these arrangements are in pairs. I have found 24 arrangements for the name Emma when I make each 'M' different. I have proved my prediction to be right.

• Word count: 2265
9. ## EMMA's Dilemma Emma and Lucy

Let's try 5 figures: 12345 13245 12354 13254 12435 --- 6 arrangements 13452 --- 6 arrangements 12453 13425 12534 13524 12543 13542 14235 15432 14253 15423 14352 --- 6 arrangements 15324 ---- 6 arrangements 14523 15342 14532 15243 14325 15234 Don't you notice the arrangements of last 4 numbers added up, it equal 24, so if we time 24 by 5, and get 120, and 120 is the total of arrangements. Carry on, if a number has 6 figure, then the total of arrangement should be 120 times by 6, and get 720, and 720 is the total of arrangements.

• Word count: 2360
10. ## permutations & combinations

KAET 7. ATEK 8. ATKE 9. AKTE 10. AKET 11. AEKT 12. AETK 13. TEAK 14. TEKA 15. TKEA 16. TKAE 17. TAKE 18. TAEK 19. ETAK 20. ETKA 21. EKTA 22. EKAT 23. EAKT 24. EATK A three-letter word will have 6 variations. This can be proved by the following word TOM 1. TOM 2. TMO 3. OMT 4. OTM 5. MOT 6. MTO A five-letter word will comprise of 120 variations. The following word will prove this: TEAMS 1. TEAMS 2. TEMAS 3. TESMA 4. TESAM 5. TEMSA 6. TEASM 7. TAEMS 8. TAESM 9. TAMSE 10.

• Word count: 1682
11. ## Math exam

2 foreldre kj�rer kanoene opp med henger. Hvor mye kostet bensinen de bruker p� � kj�re? Mat: Frokost og kvelds--> Br�d med div. p�legg. Juice, appelsin epple. Middag dag 1: Kj�ttdeig og ris. 4 stykker pr. gruppe, vi deler ut maten. Middag dag 2: spagetti p� boks. 2 bokser pr. gruppe? Middag dag 3: p�lse og lompe p� b�l, maks 3 p�lser hver. Drikke: Appelsin og eplejuice, Husk: Mer verdig avgift!! Sykkeltur: Fra Skedsmokorset til Stor�yungen.

• Word count: 541
12. ## Emma's Dilemma

When the arrangements starting with L had run out altogether, I took the next letter i.e. U and started to look for arrangements starting with that. In the end, I found that there were 24 different arrangements for the letters in Lucy's name: LUCY ULCY CLUY YLUC LUYC ULYC CLYU YLCU LCUY UCLY CULY YULC LCYU UCYL CUYL YUCL LYUC UYLC CYLU YCLU LYCU UYCL CYUL YCUL To ensure there were no errors in the results, I did the same with a similar name "Alex" and compared my findings. I found that there were 24 different arrangements for Alex's name which means that the results above should be credible: ALEX LAEX EALX XALE ALXE LAXE EAXL

• Word count: 2380
13. ## In this piece of coursework I will investigate how many times and ways I can arrange Emma's name. I will start by looking at Emma's name. Further then I will look at letters with single, double, triple and etc letters on them

I also noticed that in her name there are 6 possibilities beginning with each letter different . For example there are 6 arrangements with Lucy beginning with l and 6 beginning with u and so on. I also observed that Emma's different combinations were half of Lucy's. 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. I predict that for a five letter word there will be 120 arrangements.

• Word count: 1099
14. ## Emma's Dilemma Question One: Investigate the number of different arrangements of the letters

Question Two: Emma has a friend named Lucy, Investigate the number of different arrangements of the letters of LUCY's name. Answer: In order for me to answer this question, I will write down all of the different arrangements for the letters of Lucy's name. This will allow me to get the total number of arrangements, which will help me to find a rule in the latter questions in this piece of coursework. Below are all of the different arrangements of the letters of Lucy's name: LUCY LUYC LCUY LCYU LYUC LYCU ULCY ULYC Total: UCLY 24 UCYL UYCL UYLC CLUY

• Word count: 5904
15. ## Emma's Dilemma

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 name. I will just have to find out how many different arrangements there are. The problem with this method is that if I had 1 letter or more repeated then this method would not work.

• Word count: 1379
16. ## From Emma's name, a four letter word, we this time only get twelve arrangements. This exactly half of twenty-four,(twenty-four being the total number of arrangements for a four letter word with all the letters differ

And that continues and work, but only when all the letters in the word are different. This is done because if you have a 3 letter word to start with you have 3 letters to choose from so if I use the word THE. To start with I could use T, H or E. Then after I have used 1 of the letters I have a choice of 2 letters to use so if I start with the H I have the choice of using the T or the E and the say I use the T I only have I choice left.

• Word count: 1200
17. ## Johannes Gensfleisch zur Laden zum Gutenberg, or commonly know as Johann Gutenberg, was the inventor of the printing press. He was born in Mainz around 1397 and lived until around 1468. He is most known for his inventing of a movable type

So, around 1430, he went to live in Strassburg, where he remained until around 1444. (Johann Gutenberg 1) In Strasbourg, he joined a Goldsmith's guild, where he taught various crafts, such as gem polishing, the manufacturing of looking glasses, and the art of printing. (Johannes Gutenberg I) He worked with friends and taught them his secret profession of printing, eventually establishing his own, new better way of printing, for which he is most famous. Johannes Gutenberg invented a letterpress, in which there was a mold for each letter or character. (Johannes Guttenberg II) He invented up to three hundred, which was representative of those found on an ornate scroll or handwritten letters. (Johannes Gutenberg I)

• Word count: 1113
18. ## I have been given a problem entitled 'Emma's Dilemma' and I was given the following information: 'Emma and Lucy are playing with arrangements of their names

arrangements in which the letters can be rearranged, as shown below: ABC ACB BAC BCA CAB CBA Four Different Letters: As we were shown from the word Lucy, for a word with four different letters, there are 24 different arrangements in which the letters can be arranged, as shown below: ABCD ABDC ACBD ACDB ADCB ADBC BACD BADC BCDA BCAD BDAC BDCA CABD CADB CBAD CBDA CDAB CDBA DABD DACB DBAC DBCA DCAB DCBA Five Different Letters: For a word with five different letters, there are 120 arrangements in which the letters can be rearranged.

• Word count: 4266
19. ## Emma is playing with arrangements of the letters of her name to see how many combinations there are.

I then change the first letter and repeat this process again. I will now try another four-letter name with no repeated letters to see if there are always 24 different combinations. FAYE AYEF EFYA YAFE FAEY AYFE EFAY YAEF FEAY AFEY EYAF YEFA FEYA AFYE EYFA YEAF FYAE AEFY EAYF YFEA FYEA AEYF EAFY YFAE I have found that a four-letter name with one repeated letter has 12 combinations and a four-letter with no repeated letters has 24 arrangements. I will now try a name with two repeated letters to see how many combinations there are.

• Word count: 1977
20. ## EMMA'S DILEMMA

E.g. CA is one and AC is another one. In a three- lettered word, each letter has two arrangements. E.g. CAN and CNA are two arrangements. In a four-lettered word, each letter has 6 arrangements. E.g. LUCY, LUYC, LCUY, LCYU, LYCU and LYUC are 6 arrangements. I have also noticed that 6*4=24 and that gives the number of arrangement for a four-lettered word. 2*3=6 this gives the arrangement for a 3 lettered word, and 1*2=2 gives the arrangement for a 2-lettered word. I think there is a pattern to this because if I want get the arrangement for a 2 lettered word, then I have to get the arrangement

• Word count: 3021
21. ## Emma's dilemma The different ways of arranging letters for Emma's name

rule I can predict the 5 letters will have 120 different combinations because the previous number of combinations was 24 and the number of letters is 5 so 5 x 24 =120 There is an easier and quicker way to find how many combinations there are for a certain number this is called N factorial written as N! (N stands for the number of letters you have). N factorial times all the numbers from the number or letters you have all the way to 1 for example if you had 5 letters it would be 5!=5x4x3x2x1=120.

• Word count: 933
22. ## We are investigating the number of different arrangements of letters.

We are trying to work out a formula which can calculate the number of arrangement when we look at a number. Let's list all the arrangment for 1234: 1234 4123 1243 4132 1324 --- 6 arrangment 4231 ---- 6 arrangement 1342 4213 1432 4321 1423 4312 2134 3124 2143 3143 2341 ---- 6 arrangements 3241 --- 6 arrangements 2314 3214 2413 3412 2431 3421 So if we time 6 by 4, we would get 24, and we can get the total arrangements of 24.

• Word count: 2375
23. ## Emma's Dilemma

Therefore, I have developed my own method. Method 1. 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. Starting with the first letter, rearrange the last 2 letters of the name. If the last two letters are identical rearrange the letter before it (do not rearrange identical letters).

• Word count: 2064
24. ## I will find all the different combinations of Emma's name by rearranging the letters. Following this, I will do the same with Lucy's name and compare

I will compare these with my original results and work out a formula for names with different letters. My final investigation will be with names with sets of same letters e.g. ANNA. At the end, I will have a formula for all names. I am going to begin by investigating the number of arrangements in Emma's name. Emma EMMA EMAM EAMM AMME AMEM AEMM MEAM MAEM MMEA MMAE MEMA MAME I have found 12 different combinations for Emma's name. I will now investigate the number of arrangements of Lucy's name. Lucy LUCY LUYC LCUY LCYU LYCU LYUC ULYC UYLC UCYL UCLY ULCY ULYC CYLU CYUL CULY CUYL CLYU CLUY YLCU YLUC YCUL YCLU YULC YUCL I found out that there are 24 different arrangements of Lucy's name.

• Word count: 1121
25. ## Emma's Dilemma

If I then multiplied the number six (number of arrangements) by the number four (total amount of letters in the word) I could calculate the amount of arrangements for the whole equation. Then i could get the total amount of arrangements. 4 x 6 = 24 i can use this equation to calculate the area of a rectangle. Height x width. But in this case the height was the number of arrangements with any-one letter at the beginning and the width was the total amount of letters in anyone word.

• Word count: 879
26. ## GCSE Mathematics: Emma's Dilemma

the beginning: 4123 4132 4231 4213 4312 4321 By using the above method you will definitely acquire all the different arrangements of a four letter word with different letters. Now I shall list all the possible combinations for the word "EMMA". But there is a problem! There are 2 of the same letter (M) in "EMMA". To start off with, I shall use the same method that I used with the word "LUCY". EMMA EMAM EMMA WMAM EAMM EAMM MMAE MAME MEMA MEAM MAEM MMEA MAEM MAME MMEA MMAE MEAM MEMA AMME AMEM AMME AMEM AEMM AEMM =24 Combinations Obviously

• Word count: 2883