Emma's Dilemma.

Maths Coursework- Emma’s DilemmaFor this piece of coursework I am going to investigate the number of different ways I can write a word, re-arranging the letters without having any repeats of the sequence.After I have finished my investigations I will try and use my findings to draw together a formula which I could then use to find out how many ways a word can be written for any chosen word.My initial step is to write the name ‘EMMA’ with as many different arrangements I can find.

Part 1

Next I am again going to try a 4 letter word, but this time without repeats (no 2 letters the same) in it.I predict that a 4 letter without repeats will have a lot more letter arrangements than the name EMMA which has ‘M’ repeated.Part 2- I have chosen the name ANDY.

This is double the number of ways EMMA can be written.

My first thought to get the number of arrangements is possibly by dividing a value (which I do not know at present) by the number of times a letter is repeated in a word. E.g. SARA would be Χ/2, where Χ is the unknown value and the 2 is because letter A is repeated twice.Now, I am going to try a 4 letter word, XXXY, which has X repeated 3 times. I predict there will be less re-arrangements.Part 3

My prediction was correct, the more times a letter is repeated, then the less number of possible re-arrangements there are. I shall now try a name with 5 different letters. I will use the name MANDY. I predict that there will be 120 different arrangements. I have managed to work this out by noticing that ANDY has 24 different combinations. By adding ‘M’ to make M (ANDY) it would mean the 24 different arrangements of ANDY with M in front would still all be different. As there are 5 letters there should be 24 different combinations with each letter M, ...

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My prediction was correct, the more times a letter is repeated, then the less number of possible re-arrangements there are. I shall now try a name with 5 different letters. I will use the name MANDY. I predict that there will be 120 different arrangements. I have managed to work this out by noticing that ANDY has 24 different combinations. By adding ‘M’ to make M (ANDY) it would mean the 24 different arrangements of ANDY with M in front would still all be different. As there are 5 letters there should be 24 different combinations with each letter M, A, N, D and Y at the front i.e. M(24) A(24) N(24) D(24) and Y(24). Therefore 5 x 24 = 120 arrangements.Part 4

I will now try a 3 letter word, which has no repeats. After doing this I will try another 3 letter word, which has a letter, repeated twice. I predict there will be twice as many combinations for the word without repeats.Part 5- ROB Part 6- BOB

What I predicted was correct, I got the idea from EMMA, which has a letter repeated, having half the number of combinations than the name ANDY, which has no repeats.I think there must be a link between how many repeats there are in the word and the number of arrangements.

I will now try and predict how many arrangements there are for words I have not yet tried, I will also record my results so far in the table that follows;

I have noticed that a 5 letter word with a letter repeated twice has half the number of combinations of that of a 5 letter word without repeats. A 5 letter word that has a letter repeated 3 times has a 1/3 of the number of arrangements of that of a word, which has a letter only, repeated twice.I have also noticed that the numbers of arrangements are always multiples of the number of letters in the word. This also means the numbers of letters are always factors in the number of combinations.

I have looked up through an A-Level Maths textbook to see if there is a topic covering multiplying numbers out i.e. 4 x 3 x 2 x 1. I found that this is covered by the discipline of factorials (factorials are represented as an !). I shall try to incorporate this in arriving at a general formula for predicting the number of different patterns for any given word.

My first formula is N! / R! where N is the number of letters in the word and R is the number of times a letter is repeated. i.e. for the name EMMA the formula would be 4! / 2! = 4 x 3 x 2 x 1 2 x 1 = 12✓This works so I will try it with XXXY = 4!

3! = 4 x 3 x 2 x 1 3 x 2 x 1 = 4 ✓This also works.I shall now try this formula with a name that has 2 letters repeated. i.e. ANNA = 4! 2 x 2! = 6 ✓1) ANNA

2) ANAN3) AANN

4) NNAA5) NANA6) NAANI will now try the formula with the letters XXXYY.

To use the formula for this word it word be; 5! R!this formula cannot be used for this word though. This is because my present formula does not account for words that have more than one letter repeated. I shall now try a new formula which will account for words that have more than one letter repeated. the new formula is N! n1! n2! n 3!……………… Where n1! is the number of times the first letter repeated and n2! is the number of times the second letter is repeated and when n3! Is the number of times the third letter is repeated. The ……………… means more nx! which can be added depending on how many letters are repeated. I will try the new formula with XXXYY. = 5! 3! x 2!= 10 ✓Conclusion

My new formula, works for any given word. The N! stands for the factorial of number of letters in the word. n1! is the number of times the first letter repeated and n2! is the number of times the second letter is repeated and so on. The ………… represents more nx! which can be added depending on how many letters are repeated. I could use my formula to work out any word and my last example is going to be FINISHED.

= 8! 2! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 2 x 1 = 20160 arrangements