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  • Level: GCSE
  • Subject: Maths
  • Word count: 5964

Emma’s Dilemma.

Extracts from this document...

Introduction

Emma's Dilemma. Introduction The aim of this coursework is to investigate into the problem of different arrangements of a set of letters. By using the given guidelines and by widening the breadth of the operation I aim to find a rule or formulae that can represent and explain the situation. The basic parameters of the project are as follows: "Emma is playing with arrangements of the letters of her name. One arrangement is EMMA, another is MEAM, and another is AEMM. Investigate the number of different arrangements of the letters of Emma's name." This is the first task, which is then followed up by: "Emma has a friend named Lucy. Investigate the number of different arrangements of Lucy's name." After these two tasks I can then lead the investigation in my own direction: "Choose some different names. Investigate the number of different arrangements of the letters of the names you have chosen." An extension to the investigation is as follows: "A number of Xs and a number of Ys are written in a row such as 'XX . . . XXYY . . . Y . .' Investigate the number of different arrangements of the letters." The last section will allow me even more freedom to change variables to find more complex solutions covering a greater variety of situations. Finding The mathematical term for this type of function where a set of values is rearranged to give the maximum number of possible combinations are called permutations. I will use this term through the rest of the investigation too. Part One: I will begin by manually working out the possible permutations of the letters in the name 'Emma'. To do this I have to distinguish between each character in the name. There are four characters ??E, M, M and A. However, there are two Ms and I must be aware of this repeating letter. ...read more.

Middle

I know that that formula is going to be different because for 'Emma' there are 12 combinations, whereas when the existing formula is applied then there should be 24 combinations. With one repeating letter... One Letter: 'M' As before, with one letter there will only by one possible arrangement, regardless of how many letters are repeating in the name, because with a one letter name you cannot have repeating letters. Therefore one-letter names are not affected by repeating letters. Two Letters: 'BB' When there is one repeating letter in the name then a two-letter name acts in the same way as a one letter name; it only has one possible arrangement: 'BB', because the two Bs are indistinguishable and would create the same combination if they were swapped around. Three Letters: 'ANN' A = 1 N = 2 Combination # Combination Letter Representation 1 1 2 2 A N N 2 2 1 2 N A N 3 2 2 1 N N A A three-letter name has got three possible permutations. Four Letters: 'EMMA' As tested in part one, I already know that a four-letter name with one repeating letter has got twelve possible permutations. Five Letters: 'LINNY' L = 1 I = 2 N = 3 Y = 4 Combo # Combination Letter Representation 1 1 2 3 3 4 L I N N Y 2 1 2 3 4 3 L I N Y N 3 1 2 4 3 3 L I Y N N 4 1 3 2 3 4 L N I N Y 5 1 3 2 4 3 L N I Y N 6 1 3 3 2 4 L N N I Y 7 1 3 3 4 2 L N N Y I 8 1 3 4 2 3 L N Y I N 9 1 3 4 3 2 L N Y N I 10 1 4 2 3 3 L Y I N N Combo # Combination Letter Representation 11 1 4 ...read more.

Conclusion

This kind of occurrence, where two events together gives a smaller value, reminds me of probabilities. I.e. the probability of a coin being flipped twice and both flips landing on heads would be 0.5 x 0.5 = 0.25. In this case XXYZ gives half (0.5) the permutations of a 4-letter name with no repeats; WXYY also only gives half (0.5). Combining them together like a probability would give 0.5 x 0.5 = 0.25 the number of permutations of a 4-letter name with no repeats (24 x 0.25) = 6. Constructing a formula to show this may go as follows: P = n ! x ([n ! ? r+1 !] ? n !) x ([n ! ? d+1 !] ? n !) (where d is the number of times the second letter is repeated) This looks very complex but it should work. E.g. 1: 'ABCD' P = 4 ! x ([4 ! ? 1 !] ? 4 !) x ([4 ! ? 1 !] ? 4 !) = 24 x 1 x 1 = 24 E.g. 2: 'AABCD' P = 5 ! x ([5 ! ? 2 !] ? 5 !) x ([5 ! ? 1 !] ? 5 !) = 120 x 0.5 x 1 = 60 E.g. 3: 'AABBBC' P = 6 ! x ([6 ! ? 2 !] ? 6!) x ([6 ! ? 3 !] ? 6 !) = 720 x 0.5 x 0.167 = 60 E.g. 4: 'XXXYY' P = 5 ! x ([5! ? 3 !] ? 5 !) x ([5 ! ? 2 !] ? 5 !) = 120 x 0.167 x 0.5 = 10 All of the above examples use the formula and all the values given are correct so this formula is viable for any name with up to two different letters duplicated. If you have a name with more than two different letters duplicated then you would probably have to add additional terms to the formula to incorporate those duplicated letters into the equation. By Milan Shah - Form 11P - Set 01 Page 1 Mathematics Coursework: Emma's Dilemma Mathematics Coursework: Emma's Dilemma ...read more.

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