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# GCSE: Emma's Dilemma

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1. ## Lucy's Dilemma

I tried with other names and found there to be a pattern. I used Ben, Henry, and Thomas. The numbers were: For Ben, 6 or 3!; For Henry, 120 or 5!; and for Thomas, 720 or 6!. I realised the formula for this was: n! This is because, e.g. for 5, there are 5 possibilities for the first column, 4 for the second, 3 for the third, 2 for the forth, and 1 for the fifth. Then Lucy played around with her friend, Emma's, name. She writes down the different combinations: 1. EMMA 2. EMAM 3. EAMM 4. MEMA 5. MEAM 6. MMEA 7. MMAE 8. MAEM 9. MAME 10. AEMM 11.

• Word count: 501
2. ## Emma&#146;s Dilemma

LUCY LUYC LYUC LYCU LCUY LCYU YULC YCUL YLCU YUCL YCLU YLUC CLUY CLYU CYLU CYUL CUYL CULY UCLY UCYL UYLC UYCL ULYC ULCY As you can see there are 24 combinations. From now on in my investigation I will be using combinations of letters, which are not necessarily names, as it is easier and saves time. AB BA Letters: AB, 2 Letters. Combinations = 2 Letters: ABC, 3 letters. ABC ACB BCA BAC CAB CBA Combinations = 6. Letters: ABCDE, 5 letters.

• Word count: 1403
3. ## Emma's Dilemma

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 variations) I will use my Name which has 5 letters I have chosen not to use my name because it long and it hurts my head just trying.

• Word count: 752
4. ## Dave's Dilemma

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

1- lucy 2- luyc 3- lcuy 4- lcyu 5- lycu 6- lyuc 7- ulcy 8- ulyc 9- ucyl 10- ucly 11- uycl 12- uylc 13- culy 14- cuyl 15- clyu 16- cluy 17- cyul 18- cylu 19- yclu 20- ycul 21- yluc 22- ylcu 23- yucl 24- yulc Here are the 24 different arrangements for Lucy's name, but while she was looking over the different arrangements she noticed something strange. Even though her name and Lucy's name had the same amount of letters, Lucy's name had twice as many arrangements.

• Word count: 1309
6. ## Emma's Dilemma

1.lucy 2.luyc 3.lcyu 4.lycu 5.lcyu 6.lyuc 7.ulcy 8.ucly. 9.uylc 10. ulyc 11.uycl 12. ucyl 13.yucl 14.ycul 15.yluc 16. ylcu 17.yclu 18.yulc 19.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 if three letters were used? For example, Ted. Ted: 3 letters, none the same. 1. Ted 2.

• Word count: 1149
7. ## Emma&#146;s Dilemma

This is half the original value as some of the permutations repeat themselves. This is half of the result if all the letters are different. This could be written as (3*2*1)/(2*1) which gives the answer 3. 2 divide it, as there are 2 letters the same. Three letters, all the same. This time I have substituted the A for an M. 1. MMM There is only 1 possible permutation. This is the original amount 6/ 6, this could also be written as (3*2*1)/(3*2*1) as there are 3 letters the same in a 3-lettered word.

• Word count: 2226
8. ## GCSE Mathematics Coursework - Emma's Dilemma

The number of arrangements of letters in a 3-letter name without repeated letters is 6, and the number of arrangements in a 2-letter name without repeated letters is 2. There is only one possible arrangement of letters in a 1-letter word. The number of arrangements of letters in each name increases like this: 1x1=1 1x2=2 1x2x3=6 1x2x3x4=24 Or 1 2 6 24 X2 x3 x4 This can also be written as factorials, e.g. 2 factorial (2!)=1x2=2. 3 factorial (3!)=1x2x3=6. 4!

• Word count: 1184
9. ## Emma's Dilemma

IKRYC 37) IYRKC 38) IYRCK 39) IYCRK 40) IYCKR 41) IYKRC 42) IYKCR 43) IRCKY 44) IRCYK 45) IRKCY 46) IRKYC 47) IRYCK 48) IRYKC 49) CKYRI 50) CKYIR 51) CKRIY 52) CKRYI 53) CKIRY 54) CKIYR 55) CYRIK 56) CYRKI 57) CYIRK 58) CYIKR 59) CYKRI 60) CYKIR 61) CRYKI 62) CRYIK 63) CRKYI 64) CRKIY 65) CRIKY 66) CRIYK 67) CIRKY 68) CIRYK 69) CIKYR 70) CIKRY 71) CIYKR 72) CIYRK 73) KYRIC 74) KYRCI 75) KYICR 76) KYIRC 77) KYCRI 78) KYCIR 79) KRCIY 80) KRCYI 81) KRIYC 82) KRICY 83) KRYIC 84) KRYCI 85) KICYR 86) KICRY 87) KIYCR 88) KIYRC 89) KIRYC 90) KIRCY 91) KCYIR 92) KCYRI 93) KCIYR 94) KCIRY 95)

• Word count: 1357
10. ## Emma&#146;s Dilemma

Table of Results: - A pattern can be recognised from this table. 1 x 2 = 2 1 x 2 x 3 = 6 1 x 2 x 3 x 4 = 24 All you need to do is multiply the top numbers to get the number of combinations. This is called "Factorial Notation" and there is a much simpler way to write this. You just simply put an "!" after the number that is to be multiplied up to.

• Word count: 2746
11. ## Emma's Dilemma

There are 24 different possibilities in this arrangement of 4 letters all different. Double the amount as before with Emma's name, which has 4 letters and 3 different. I have noticed that with Lucy there are 6 possibilities beginning with each different letter. For example 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. I will choose some different names. JO OJ There are 2 different letters in this name and there are 2 different arrangements.

• Word count: 1052
12. ## GCSE Maths Coursework : Emma&#146;s Dilemma

abcd abdc acbd acdb Instead of doing the rest of my investigation with names I decided to use letters i.e. abcd, to complete my investigation as I think it would be even easier to arrange them systematically and spot patterns.

• Word count: 331
13. ## Emma&#146;s Dilemma

The ways are:- 1.lucy 9.uycl 17.clyu 2.lcuy 10.ulyc 18.cluy 3.lcyu 11.ucyl 19.yluc 4.luyc 12.uylc 20.ylcu 5.lyuc 13.cyul 21.yucl 6.lycu 14.cylu 22.yulc 7.ulcy 15.culy 23.yclu 8.ucly 16.cuyl 24.ycul I am now going to use Ann a 3 letter word with 2 letters the same I will get 3 different ways.

• Word count: 560
14. ## Emma&#146;s Dilemma

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

• Word count: 1354