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  • Level: GCSE
  • Subject: Maths
  • Word count: 1309

Emma's Dilemma

Extracts from this document...

Introduction

Emma is playing with the different arrangements of the letters in her name, here is a list of all the different arrangements: 1- emma 2- emam 3- eamm 4- amme 5- amem 6- aemm 7- mmae 8- mmea 9- mema 10- mame 11- maem 12- meam To make sure that she listed all of the possible arrangements, Emma used a systematic formula. Lets try it with the letters abcd: 1-abcd first start off by keeping the first two letters the same and then swapping the last two letters. 2-abdc Then keep the first letter where it is and swap all the second number with the third. 3-acbd Now you can swap the last two numbers and keep the first two where they are. 4-acdb Next swap the second letter with the only letter that has not been second yet which should be D, then you can swap the last two letters again. 5-adbc 6-abdc Now that you have listed all of the arrangements for the letter A you can repeat this process but each time swap the first letter for one of the others. 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 Emma then tried to find the different arrangements of her friend Lucy's name, she used the same systamatic formula. ...read more.

Middle

hcad 16- hcda 17- hdac 18- hdca 19- dcha 20- dcah 21- dhac 22- dhca 23- dach 24- dahc This name had 24 different arrangements. She then created a table. No of letters No of arrangements 3 6 4 24 5 120 She then realised that to work out the number of arrangements you would have to times the amount of letters in that name by the number of arrangements of a name with 1 less letter. So for example to work out how many arrangements a name with 6 letters had you could times 6 by 120 and you should get 720. Emma then listed the arrangements of a name with six different letters and it worked. She then continued the table. 6 720 7 5040 8 40320 9 362880 Emma thought about it for a second and realised another way of working out the number of arrangements in a name with all different letters. If the name had 6 letters then you could try the following formula: 1*2*3*4*5*6= 720 She then tried this for the others from the table and it worked. 7fig- 1*2*3*4*5*6*7= 5040 8fig- 1*2*3*4*5*6*7*8= 40320 9fig- 1*2*3*4*5*6*7*8*9= 362880 On her calculator Emma could work out this formula quickly by pressing on button, below is ...read more.

Conclusion

She divided it by two because there were 2 of the same figure. Emma then thought about how she would work out the number of arrangements for a name with more than two of the same figures or two sets of the same figure. She tried to use the same formula as for a name with 2 of the same figures. She listed some various names, which had more than one set of the same figure. Terence Ponter 13 letters 2 T's 4 E's 2 N's 2 R's She thought about the name and decided to use trial and improvement with some different methods. First she decided to enter the calculation below into her calculator. 13! = 32'432'400 (2!*4!*2!*2!) She tried this method for some other names. Chris Aust 9 letters 2 S's 9! = 181'440 2! Daniel Assiter 13 letters 2 A's 2 E's 2 I's 2 S's 13! = 389'188'800 (2!*2!*2!*2!) By looking at the results from these names, Emma was able to work out the following formula. N=number of arrangements X= number of letters A, B, C, D= figure which appears more than once in the name (each group of same letters is shown by a different letter) ...read more.

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