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• Level: GCSE
• Subject: Maths
• Word count: 1052

# Emma's Dilemma

Extracts from this document...

Introduction

1 Introduction and Planning The problem set involves investigating the number of different arrangements for different amounts of letters. One arrangement is: EMMA A different arrangement is: EAMM Another arrangement is: MEAM I am going to investigate the number of different arrangements for the letters in the name EMMA. The name EMMA consists of 2 different letters and 2 letters the same. I am going to investigate other names with the same combination of letters; I will try and find a formula for these combinations I will then look at the name LUCY. This combination consists of all the letters being different. I will also try and find a formula for this arrangement, and number of letters. EMMA, AMME, AMEM, EMAM, AEMM, EAMM, MMEA, MMAE, MEMA, MAME, MEAM, MAEM. There are 12 possibilities; note that there are 4 total letters and 3 different. What if they were all different like Lucy? Arrangements for Lucy: 2 Lucy, ucyl, cylu, ycul, Luyc, ucly, cyul, yclu, Lcuy, ulcy, culy, yulc, Lcyu, ulyc, cuyl, yucl, Lyuc, uycl, clyu, ylcu, Lycu, uylc, cluy, and yluc. ...read more.

Middle

Taking for example a 3-letter word, I have worked out that if we do 3 (the length of the word) 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 which is the same as 4 x 3 x 2. So, by using factorial (!) I can predict that there will be 5040 different arrangements for an 7-letter word. The formula for this is: n! = A Where n = the number of letters in the word and A = the number of different arrangements. I want to use another method to convey my work and understanding. qlucy qucyl qcylu qycul qluyc qucly qcyul qyclu qlcuy qulcy qculy qyulc qlcyu qulyc qcuyl qyucl qlyuc quycl qclyu qylcu qlycu quylc qcluy qyluc I can see that the numbers for different arrangements are going to increase as more different letters are used. ...read more.

Conclusion

Also in a 5-letter word there are 120 arrangements and 24 arrangements (120 divided by 5) for each letter. As there IS a divide issue involved I had a go at trying to work out a formula. I came up with; The number of total letters factorial, divided by the number of x's, y's ect factorised and multiplied. Formula is n!/x!y! For example: A five letter word like aaabb; this has 3 a's and 2 b's (3 x's and 2 y's). So: 1x2x3x4x5 / 1x2x3 x 1x2 = 120 / 12 = 10!!! A four letter word like aabb; this has 2 a's and 2 b's (2 x's and 2 y's) So: 1x2x3x4 / 1x2 x 1x2 = 24 / 4 = 6!!! A five letter word like aaaab; this has 4 a's and 1 b (4 x's and 1 y) So: 1x2x3x4x5 / 1x2x3x4 x 1 = 120 / 24 = 5!!! Five letter words like abcde; this has 1 of each letter (no letters the same) So: 1x2x3x4 / 1x1x1x1x1x1 = 24 / 1 = 24 All these have been proved in previous arrangements. This shows that my formula works!!! ...read more.

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1. ## Arrangements for names.

As each letter has its own number of arrangements i.e. there were 5 beginning with a, and 5 beginning with b, I think that factorial has to be used again. Also in a 5-letter word there are 120 arrangements with 24 arrangements (120 divided by 5)

2. ## Emma's Dilemma

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