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GCSE: Emma's Dilemma

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1. Emma's Dilemma.

LUCY, LUYC, LYCU, LYUC, LCYU, LCUY, ULCY, UCLY, UYLC, ULYC, UCYL, UYCL, YCUL, YUCL, YULC, YLCU, YLUC, YCLU, CYLU, CYUL, CUYL, CULY, CLUY, CLYU. Here as you can see there are 24 different ways of arranging the name Lucy. A quicker way of writing this down would have been to see how many arrangements you could make by leaving L as the first letter. There are 6 when structured like this. After you have all the L ways are worked out, then just times that number (6)

• Word count: 1070
2. Emma&#146;s Dilemma.

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) LYUCE 19) LCYUE 2) LUCYE 8) LEUYC 14) LYUEC 20) LCYEU 3) LUYCE 9) LEYCU 15) LYCUE 21) LCEYU 4) LUYEC 10) LEYUC 16) LYCEU 22) LCEUY 5) LUEYC 11) LECUY 17) LYEUC 23) LCUYE 6) LUECY 12) LECYU 18)

• Word count: 1322
3. To investigate the number of different arrangements of letters in a different words

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. I found this out in the school library which is the same as 4 x 3 x 2 but times 1 but this make no difference to the result. This means that the formula for the number of arrangements for a word with no repeated letters is: n!

• Word count: 1200
4. Investigate the number of arrangements of Dave's name.........

Number of letters Number of arrangements Calculation 2 2 1x2 3 6 1x2x3 4 24 1x2x3x4 5 120 1x2x3x4x5 6 720 1x2x3x4x5x6 7 5040 1x2x3x4x5x6x7 From this I have concluded that the formula n! is correct( n representing the number of letters). Therefore, to find the number of arrangements for six letter word, you would multiply the number of letters (6) by the number of arrangements of the previous number (120). This gives seven hundred and twenty arrangements. I then tried to simplify this.

• Word count: 1420
5. Emma's Dilemma.

combinations 5 Letters different ANDIE ANDEI ANEDI ANEID ANIDE ANIED ADNIE ADNEI ADENI ADEIN ADIEN ADINE AIDNE AIDEN AIEDN AIEND AINED AINDE AEIDN AEIND AENID AENDI AEDNI AEDIN So far I have 24 combinations. All combinations begin with 'a' . As previous results have shown the combinations beginning with the following letters. Either e d n or I will each give 24. Therefore for the total number of combinations all I need to do is multiply 24 by 5. This gives a total of 120 different combinations for the five-letter name, with all letters different 120.

• Word count: 985
6. Compare and contrast the fictional letters in 'Birdsong' with the real letters written to Vera Brittain by Edward Brittain and Roland Leighton. Which do you find more moving and powerful and why?

In 'Weir's' letter the first thing he says is "We are going to attack." As readers we know how horrific the battle was, so immediately we can sense the naivety and the blindness of his character. In all the letters, there is a constant repetition of this. 'Stephen' is the only character who can see the premonition that "some crime against nature is about to be committed." His letter is the only one, which shows negativity towards the war. He is sending his letter to 'Isabelle,' fully aware that she may not receive it.

• Word count: 1506
7. Emma's Dilemma

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, OJ 2 letters - 2 same = 1 DD 3 letters - 0 same = 6

• Word count: 1064
8. Emma's dilemma.

* CAINE CAEIN CIEAN CNEIA CENIA * CAIEN CIANE CIENA CNIEA CENAI * CANEI CIAEN CNAIE CNIAE CEIAN * CANIE CINEA CNAEI CEAIN CEINA * CAENI CINAE CNEAI CEANI .............. The above arrangements show the first 24 arrangements for the five-letter name CAINE. The first 24 arrangements all Have the letter C in the front and in this same way if I was to carry on every other letter would come to the front and therefore there would be 24 x 5 arrangements, which is 120.

• Word count: 1333
9. Emma's Dilemma

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 CBA ACB BCA CAB There are six permutations with three different letters. d) With four letters. * 5 ABCD BACD CABD DABC ABDC BADC CADB DACB ACBD BCAD CBAC DBAC ACDB BCDA CBCA DBCA ADCB BDAC CDAB DCAB ADBC BDCA CDBA DCBA There are twenty-four permutations with four different letters.

• Word count: 1206
10. I am doing an investigation into words and their number of combinations. I will find formulae and work out the number of combinations for the words.

5 different letters 120 1x2x3x4x5=120 or 5! =120 I will now prove that 5 different letters has 120 combinations: ABCDE=24 ABCED=24 ABECD=24 AEBCD=24 EABCD=24 Total=120 I worked out the formula was n! after first discovering the pattern was 1x2; 1x2x3; as so on. From prior knowledge I knew an easier way to write this was n! or n factorial. The factorial pattern is: 1!=0x1=1 2!=0x1x2=2 3!=0x1x2x3=6 4!=0x1x2x3x4=24 Now I will draw a graph to show that I have got the right formula and to see the relationship between the number of combinations and the number of letters even more clearly than in a table.

• Word count: 3921
11. Emma's Dilemma

LUCY LCUY LCYU LUYC LYCU LYUC SIMON SIMNO SIOMN SIONM SINOM SINMO SMION SMOIN SMINO SMONI SMNOI SMNIO SOIMN SOINM SOMIN SOMNI SONIM SONMI SNIMO SNIOM SNMIO SNMOI SNOIM SNOMI (There are 24 combinations, so taking into account that this number would be possible starting with each letter there will be 24*5 combinations which equals 120 combinations) Here is a table containing all of the results that I have recorded. Hopefully when I have rounded them all together I might be able to work out a recurring pattern.

• Word count: 2145
12. Emma's Dilemma

Sam Sam Sma Ams Asm Mas Msa Sam is a 3 letter word, and all the letters are different, giving us a total of 6 combinations. JJ JJ JJ is a letter name and both of the letters are the same; therefore there is only 1 combination of this name. Jo Jo Oj Jo is a 2 letter name and both of the letters are the same giving us a total of 2 combinations. No. of letters All different 2 Same 3 Same 4 same 5 Same 2 letters 2 1 3 letters 6 3 1 4 letters 24 12

• Word count: 640
13. Emma's Dilemma

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.

• Word count: 425
14. Emma's Dilemma

This is because I have noticed something as I was working out the ?. I have found that I was going to repeat everything I was doing 4 times. Similarly, I have realised that I was doing every thing 3 times for the second letter, twice for my third letter and once for my last. Hence, I could say that to work out the ? for a four letter word, I am merely multiplying 4*3*2*1 and this gives me 24, which is the correct answer. I could express this as 4!. Could I however say this for words made up of a different amount of letters?

• Word count: 1609
15. To investigate the combination of arrangement of letters in Jeans name and then for her friend Emma.

E A M A M M E A M E M A E M M My Prediction was incorrect because although there are 4 letters in Emmas name 2 of them are the same. So there are less combinations. Now I will try some names of my own to check the combinations are always the same for names like these. POLO: I predict there will be 12 combinations because 2 letters are the same. P O L O P O O L P L O O L O O P L O P O L P O O O O

• Word count: 2289
16. Emma's Dilemma

ABC BCA There are 6 combinations for 3 letters that ACB CAB are all different. BAC CBA ABCD CABD There are 24 combinations for 4 letters ABDC CADB that are all different. ACBD CBAD ACDB CBDA ADBC CDAB ADCB CDBA BACD DABC BADC DACB BCAD DBAC BCDA DBCA BDAC DCAB BDCA DCBA From the previous information, I can form this table which may later help me in finding a formula: Table 1: Number of letters (n) Number of combinations (c)

• Word count: 1307
17. Emma's Dilemma

Table of Results Number of Letters Number of Different Arrangements 2 2 3 6 4 24 From the first two results the amount of different arrangements possible appears to be double the number of letters in the word. Looking at the results of the 4-letter word it appears that this does not apply to them all. Therefore I believe it may have something to do with the number of letters but, be higher as the number of letters increases rather than just doubling.

• Word count: 1349
18. Emma's Dilemma

I.E. The possible arrangements for a 5 letter name with no repeats is (24x5) I shall now test my hypothesis: JAMES JAEMS JASEM JASME JAMSE JAESM JSAME JSMEA JSMAE JSAEM JSEAM JSEMA JMASE JMAES JMESA JMEAS JMSAE JMSEA JEAMS JEASM JESAM JESMA As there is five letters in the name James there will obviously be 120 arrangements as there is 24 for each letter JEMAS JEMSA Just as I had predicted there are a possible 120 arrangements for a five letter name with no repeats. Now I have proved my original hypothesis I shall now do a table of results: Table of Results Number of Letters Number of Different Arrangements 2 2 3 6

• Word count: 1833
19. Emma's Dilemma

Investigate the number of the different arrangements of the letter s of LUCY 's name. 1) EHSAN EAHNS EHANS ENAHS ESAHN EAHSN EHASN ENASH ESANH EANHS EHNAS ENHAS ESHNA 6x4=24 EANSH EHNSA ENHSA ESHAN There is 24 different EASHN EHSAN ENSAH ESNAH arrangements for EASNH EHSNA ENSHA ESNHA letter E. HAENS HEANS HNAES HSAEN HAESN HEASN HNASE HSANE 6x4=24 HANES HENAS HNEAS HSEAN There is 24 different HANSE HENSA HNESA HSENA arrangements for HASEN HESAN HNSAE HSNAE letter H.

• Word count: 1443
20. Emma's Dilemma - Rearranging Emma's Name in different permutations

That's twice as many as EMMA, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For instance there are 6 arrangements with Lucy beginning with L, and 6 beginning with u and so on. 6 X 4 (the amount of letters) gives 24. From these 2 investigation I worked out a method: I used numbers to represent the letters, Step1: 1234---Do the last two number first then you get 1243. 1243---Do the last three numbers and try the possibility. 1423.

• Word count: 1249
21. Emma's Dilemma

IAEKT EAIKT KTEIA ATEIK TAEIK IAETK EAITK KIATE AIKTE TIKAE ITKAE ETKAI KIAET AIKET TIKEA ITKEA ETKIA KITAE AITKE TIAKE ITAKE ETAKI KITEA AITEK TIAEK ITAEK ETAIK KIEAT AIEKT TIEKA ITEKA ETIKA KIETA AIETK TIEAK ITEAK ETIAK KEATI AEKTI TEKAI IEKAT EIKAT KEAIT AEKIT TEKIA IEKTA EIKTA KETAI AETKI TEAKI IEAKT EIAKT KETIA AETIK TEAIK IEATK EIATK KEIAT AEIKT TEIKA IETKA EITKA KEITA AEITK TEIAK IETAK EITAK There are 120 different arrangements. TABLE OF RESULTS Number of letters (n) Number of different letters Number of different arrangements (a)

• Word count: 4324
22. Emma's Dilemma

1)AT 2)TA I will now be investigating the number of different arrangements of the letters in DAVIS's name 1)DAVIS 2)DAVSI 3)DAISV 4)DAIVS 5)DASIV 6)DASVI 7)DVAIS 8)DVASI 9)DVISA 10)DVIAS 11)DVSAI 12)DVSIA 13)DSVIA 14)DSVAI 15)DSAVI 16)DSAIV 17)DSIAV 18)DSIVA 19)DIAVS 20)DIASV 21)DIVAS 22)DIVSA 23)DISVA 24)DISAV Looking at my results I have created a results on the information. 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 letter word has 2 arrangements, and a 3 letter word has 6.

• Word count: 1325
23. Emma's Dilemma

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.

• Word count: 781
24. Emma's Dilemma

LUCY UYCL YCLU LCUY UYLC YULC LCYU CULY YLCU LYCU CUYL YLUC LYUC CLYU LUYC CLUY ULCY CYLU UCLY CYUL UCYL YUCL ULYC YCUL From this second experiment I have found out that when the letter "L" is at the front of the name Lucy then there are six different combinations. The same rule applies to the letters "U", "C" and "Y" in the name Lucy. Overall there are 24 different combinations for the name Lucy From the first two experiments I have noticed that the name Emma has half the number of combinations when compared with the name Lucy.

• Word count: 850
25. Emma's Dilemma

I chose 3 names under each amount of letters. 2 Letter Names This is all the different arrangements of the name JO JO OJ This is all the different arrangements of the name MO MO OM This is all the different arrangements of the name TY TY YT 3 Letter names This is all the different arrangements of the name JIM JIM JMI MIJ IMJ IJM MJI This is all the different arrangements of the name BEN BEN NEB BNE ENB EBN NBE This is all the different arrangements of the name ANN.

• Word count: 806