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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

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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. 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
4. 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
5. 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
6. Emmas dilemma

All the names I investigate will all have different letters. Name: Jo Total number of letters: 2 jo oj Jo has 2 different combinations. Name: Ian Total number of letters: 3 ian ina ani ain nai ain Ian has 6 different combinations. I have realised that the total number of combinations is the number of letters multiplied by the previous number of combinations. Name: Lucy Total number of letters: 4 Previous number of combinations: 6 Prediction: 4 x 6 = 24 lucy luyc lcyu lcuy lycu lyuc ulcy ulyc uylc uycl ucly ucyl cluy clyu culy cuyl cyul cylu yluc ylcu yclu ycul yucl yulc As i predicted 24 combinations.

• Word count: 755
7. Emma's Dilemma

that do not have any repeats and are one, two, three, four and five letters long: A JO OJ MAX MXA AMX AXM XMA XAM MIKE MIEK MKIE MKEI MEIK MEKI IMKE IMEK IKME IKEM IEMK IEKM KMIE KMEI KIME KIEM KEMI KEIM EMIK EMKI EIMK EIKM EKMI EKIM MINTY MNITY MTINY MYINT MINYT MNIYT MTIYN MYITN MITNY MNTIY MTNIY MYNIT MITYN MNTYI MTNYI MYNTI MIYNT MNYIT MTYIN MYTIN MIYTN MNYTI MTYNI MYTNI IMNTY INMTY ITMNY IYMNT IMNYT INMYT ITMYN IYMTN IMTNY INTMY ITNMY IYNMT IMTYN INTYM ITNYM IYNTM IMYNT INYMT ITYMN IYTMN IMYTN INYTM ITYNM IYTNM NMITY NIMTY NTMIY

• Word count: 800
8. Emma's Dilemma

Here is another example using the name TIM. Total number of letters: 3 Previous number of combinations: 2 3x2=6 Here is another example using the name DAISY Total number of letters: 5 Previous number of combinations: 24 24x5=120 After doing this I realised that I can work out the amount of combinations using factorial notation. This is when a number is multiplied by the previous consecutive numbers. For example, using the letter 6 you would do this: 6x5x4x3x2x1 I realised this is because if I could find the total number of combinations by multiplying the total number of letters by the previous number of combinations it was the same as multiplying the total number letters by its previous consecutive numbers.

• Word count: 812
9. Emma's dilemma.

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.

• Word count: 759
10. I am investigating the number of different arrangements of letters in a word.

=24 + * E (digo) =24 + * G (dieo) =24 + * O (dieg) =24 + =120 Total no. Of letters No repeated letters 1 1 2 2 3 6 4 24 5 120 In a 4-letter name such as Lucy, the way to work out the number of combinations is to do 4 x 3 x 2 x 1 or 4 factorial (4!). . Factorial is a number multiplied by the previous consecutive numbers. Factorial notation is symbolised using an exclamation mark! So I came to conclude that the formula to work out any the combinations is the numbers of letters factorial equals arrangements (l!

• Word count: 787
11. Emma&#146;s Dilemma.

Wrote down the 2 other possibilities. There were six possibilities for the word beginning with L. I repeated the process with the letters U, Y and C. I found out that there were 6 possibilities for each of these letters and then I multiplied 6 by 4 (the number of letters) which gave me 24 possibilities for the word LUCY. Part 2 Introduction In this investigation, I am going to find out the different arrangements of the letters of the name EMMA Here are the possibilities for the name EMMA EMMA EMAM EAMM MEMA MEAM MMEA MMAE MAEM MAME AMME

• Word count: 772
12. Ivy Rowe's ideas of the past in Fair and Tender Ladies.

These letters, being a window into her mind, show us the progression of her as she grows. There is one letter in particular, which shows how important this correspondence is to her. "I hate you, you do not write back nor be my Pen Friend I think you are the Ice Queen instead. I do not have a Pen Friend or any friend in the world, I have only Silvaney who laghs and laghs and Beulah who is mad now all the time and Ethel who calls a spade a spade...I will not send this letter as I remain your hateful, Ivy Rowe."(Smith, 17)

• Word count: 994
13. Investigate all the possible combinations there are of a person's name, using only the letters in their name.

From these results I know that a 4 letter name, with no repeated letters gives 24 possible combinations, my initial thoughts where that you simply times the number of letters by 6 to get the number of possible combinations. I will be investigating this is Part 3 Part 2 Emma In this section I will be exploring the possible combinations of a name that includes a double letter. Possible Combinations: 1. emma 2. emam 3. eamm 4. meam 5. maem 6.

• Word count: 775
14. In this project I had to find out the number of different ways you could arrange any letter words with no repeats, then with one repeat and then find a formula for any number of repeats.

I found these ways of arranging it: Lucy Lcuy Lcyu Luyc Lyuc Lycu Cuyl Cyul Cylu Culy Clyu Cluy Ulcy Ucly Uycl Uylc Ucyl Ulyc Ycul Yclu Yluc Ylcu Yucl Yulc I thought this answer was 24 because there were 4 different letters and 6 different ways to arrange it with the same first letter and so 6x4 is 24 The next name I did was Chloe. I came up with 120 ways of arranging it. For each different first letter there were 24 different ways of arranging the other letters, as ; Chloe Chleo Choel Cheol Cheol Chelo Cloeh

• Word count: 619
15. Emma's Dilemma.

found an equation, which will tell you the number of letter combinations in each word (except EMMA), it is based on the following idea: In the word FRED for example, there are four letters. When rearranging the letters, there are four possibilities for where the first letter could be placed, and for each of those four possibilities, there are then another three possibilities for where the remaining three letters can go. This means in total so far there are 12 (4x3)

• Word count: 910
16. Emma's Dilemma.

3 same, 4 arrangements AAAA - 4 letter word, 4 same, 1 arrangements KATIE, KATEI, KAEIT, KAETI, KAITE, KAIET, KTAIE, KTAEI, KTEIA, KTEAI, KTIAE, KTIAE, KIAET, KIATE, KIETA, KIEAT, KITEA, KITAE, KEITA, KEIAT, KEATI, KEAIT, KETIA, KETAI, AKTIE, AKTEI, AKEIT, AKETI, AKITE, AKIET, ATKIE, ATKEI, ATEIK, ATEKI, ATIKE, ATIKE, AIAKT, AIKTE, AIETK, AIEKT, AITEK, AITKE, AEITK, AEIKT, AEKTI, AEKIT, AETIK, AETKI, TAKIE, TAKEI, TAEIK, TAEKI, TAIKE, TAIEK, TKAIE, TKAEI, TKEIA, TKEAI, TKIAE, TKIAE, TIAEK, TIAKE, TIEKA, TIEAK, TIKEA, TIKAE, TEIKA, TEIAK, TEAKI, TEAIK, TEKIA, TEKAI, IATKE, IATKI, IAEKT, IAETK, IAKTE, IAKET, ITAKE, ITAEK, ITEKA, ITEAK, ITKAE, ITKAE, IKAET, IKATE, IKETA,

• Word count: 859
17. Emma&#146;S Dilemma

There are two letters the same, which should reduce the amount of combinations produced. -Emma -Meam -Aemm -Emam -Eamm -Maem -Mmae -Mmea -Amme -Amem -Mame -Mema 12 Combinations It has been difficult to produce a formula straight away because of the repeated letter 'M'. I have decided to move on to a four-letter name in which all the letters differ, LUCY. 2) LUCY -Lucy -Lcuy -Lycu -Lcyu -Lyuc -Luyc -Ulcy -Ulyc -Uycl -Uylc -Ucly -Ucyl -Clyu -Cluy -Cyul -Cylu -Culy -Cuyl -Ylcu -Yluc -Yucl -Yulc -Ycul -Yclu 24 Combinations The name LUCY proves that a word with two letters the same has fewer combinations because it has 24 combinations, where as EMMA had only 12 combinations.

• Word count: 526
18. Psychologists have identified several "laws" of perceptual organisation on grouping which illustrate their view that the perceived whole of an object is more than the sum of its parts; that objects are interpreted as "gestalten".

Similarity: Similar figures tend to be grouped together. So, the triangles and circles (right b) are seen as columns of similar shapes rather than rows of different shapes. Good Continuation: We tend to perceive smooth, continuous patterns rather than discontinuous ones. The pattern (right c) could be seen as a series of alternating semi-circles, but tends to be perceived as a wavy line and a straight line. Music and speech are perceived as continuous rather than a series of separate sounds. Closure: The law of closure says that we often supply missing information to close a figure and separate it from its background.

• Word count: 506
19. 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
20. 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
21. 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
22. 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
23. 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
24. 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
25. 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
26. 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